According to the Chomsky hierarchy, both the statements are correct.

The language generated by given grammar is - L={b,bb,abb,abbb,ababb}
Hence it is a^mbⁿ | n>m, m>=0

From the grammar, we can easily infer that it generates all the palindromes of odd length, for ex: strings {aba, bab, aaa, bbb, ….}

Grammar is belongs to Type 0 ,Type 1,Type 2,Type 3 because it is regular. Option D is right answer because they are asking only highest type number.

The required string consists of only only matched parenthesis enclosed with parentheses.
To generate the string ((() ()) ()) , we need to perform the steps sequentially.
As the string starts with left parentheses, we will start with the production S → (S).
Depending upon the string we will go for either left or right tree generation.
1) S → (S) // Replacing S by (S)
2) S → (SS) // Replacing by S by SS
3) S → ((S)S) //Replacing S by(S)
4) S → ((SS)S)// Replacing S by SS
5) S → (((S)S))S)
6) S → (((S)(S))S)
7) S → (((S)(S))(S)) [S → ε]
8) S → ((()(S))S) [S → ε]
9) S → ((()())S) [S → ε]
10) S → ((()()))

If we have no repetition in a string then the number of substrings can be found using the formula :
n*(n+1)/2 + 1
We have added 1 because it may include a NULL string also.
The number of substrings = 10*(11)/2 +1
The number of substrings = 56

Kleene closure
It is the infinite set of all possible strings of all possible lengths over Σ(input set) excluding ʎ(empty string).

→ All possible strings of length 2 over Σ = {a,b} = 4(aa,ab,ba,bb}.
→ We are computing all possible strings using 2ⁿ formula. Where n is number of strings.

The language L={ x​ ⁿ​ y^{​ n}​ , n>=1 }
For n=1, xy
n=2,x​ ^2​ y​ ²
n=3,x​ ³​ y^{​ 3}
In the above language , there are equal number of x’s followed by y’s
Definition-1:E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyy
Which is nothing but equal number of x’s followed by equal number of y’s
X​ ⁺​ means one or more x’s
Y​ ⁺​ means one or more y’s
Definition -2 :
The expression xy|x​ ⁺​ x yy​ ⁺ ​ gives xy, xxyy , xxxxyy , xxyyyyy and so on
The definition-2 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.
Definition -3 :
The expression x​ ⁺​ y​ ⁺​ gives xy, xxy,xxxy,xyy,xyyy and so on strings.
The definition-3 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.

The language L={ x​ ⁿ​ y^{​ n}​ , n>=1 }
For n=1, xy
n=2,x​ ²​ y^{​ 2}
n=3,x​ ³​ y^{​ 3}
In the above language , there are equal number of x’s followed by y’s
Definition-1​ :E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyy
Which is nothing but equal number of x’s followed by equal number of y’s
X​
+​
means one or more x’s
Y​
+
​ means one or more y’s
Definition -2 : The expression xy|x​ ⁺​ x yy^{​ +} ​ gives xy, xxyy , xxxxyy , xxyyyyy and so on
The definition-2 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.
Definition -3 :
The expression x​ ⁺​ y^{​ +}​ gives xy, xxy,xxxy,xyy,xyyy and so on strings.
The definition-3 won’t give equal number of x’s followed by y’s
It generates any number of x’s followed by any number of y’s.

We can’t generate the strings “a” and “b” because the Starting symbol is R→ XRX|S.

Option-A:​ TRUE: S is non terminal symbol, ∈ is the null string, is ambiguous.
Option-B:​ FALSE: An unambiguous grammar has same leftmost and rightmost derivation.
Option-C:​ TRUE: An ambiguous grammar can never be LR (k) for any k, because LR(k) algorithm aren’t designed to handle ambiguous grammars. It would get stuck into undecidability problem, if employed upon an ambiguous grammar, no matter how large the constant k is.
Option-D: ​ TRUE: Recursive descent parser is a top-down parser

Here, there is no terminal symbol ‘a’, so the answer never be option B,C,D.
Option-A is also wrong because the string generated by grammar is bccccd.

→ Phrase structure grammars are also known as constituency grammars. The defining trait of phrase structure grammars is thus their adherence to the constituency relation, as opposed to the dependency relation of dependency grammars.
→ Most general phase structured grammar is context sensitive grammar.

L1 is the language when n=0, m=0.
L1={λ}
L2 is the language when only n=0
L2={bc, bbcc, bbbccc,......}
L3 is the language when only m=0
L3={ ac, aacc, aaaccc, ........}
L4 is the language when n≠0 and m≠0
L4= {abcc, aabbcccc, aaabbbcccccc,...............}
L= {L1 U L2 U L3 U L4}
L= { λ, bc, bbcc, bbbccc,........, ac, aacc, aaaccc,.........., abcc, aabbcccc,aaabbbcccccc,.........}
Option A is incorrect because it does not have any stoping point for condition when m=0 i.e it can't generate strings { ac, aacc, aaaccc, ........}
Option B is incorrect because it does not have any stopping point for condition when n=0 i.e it can't generate strings {bc, bbcc, bbbccc,......}
Option C is correct because it can generate all the strings generated by the language L.
Option D is incorrect because state S1 is unreachable in it. And because of this strings {bc, bbcc, bbbccc,......{abcc, aabbcccc, aaabbbcccccc,...............} can't be generated.

C and C++ are context sensitive languages.

According to given the given language L when
m=0, n<= 3 i.e. L1= { λ, a, aa, aaa}
m=1, n<=4 i.e. L2={ ab, aab, aaab, aaaab}
m=2, n<=5 i.e. L3={abb, aabb, aaabb, aaaabb, aaaaabb} and so on.
Hence L = L1 U L2 U L3 U ...........
L= { λ, a, aa, aaa, ab, aab, aaab, aaaab, abb, aabb, aaabb, aaaabb, aaaabb, .................}
Option(A) is not correct because it can't generate λ.
Option(B) is not correct because it can't generate "a".
Option(C) is not correct because it can't generate "a".
Option(D) is correct because it is generating the language L
.

→ Phrase structure grammars are also known as constituency grammars. The defining trait of phrase structure grammars is thus their adherence to the constituency relation, as opposed to the dependency relation of dependency grammars.
→ Most general phase structured grammar is context sensitive grammar.

→ The given language is Context free language(CFL) because we can push a's onto the top of stack as a's come as input and when a "b" come as input don't push it on the top of stack just change the state and after that when c's come come as input pop one "a" from top of stack for each "c".
→ In this way the given language can be accepted by the deterministic pushdown automata and hence the language is deterministic context free language. And when a language is CFL it is also a recursive language.
Hence the correct answer is option (C)

As we know, for any regular expression R,
Rϕ = ϕR=ϕ
So L₁ L₂ * = ϕ
and L₁ * = {ϕ}* = {ϵ}
So L₁L₂* ∪ L₁* = {ϵ}

Option(A) : it is correct option because L= {aⁿ bⁿ cⁿ} is a context sensitive language and it can also be accepted by the turing machine. Hence every CSL also recursive language.

Option(B): It is incorrect option.

Option(C): Yes it is true that recursively enumerable languages are closed under union.

Option(D): It is a correct option because recursively enumerable and recursive languages are closed under reversal.

NOTE: The difference between recursive and recursively enumerable language is that if a language is recursive then there will be definitely a turing machine which will halt and will say whether the string is accepted or not. But if a language is recursively enumerable then the turing machine will halt if the string is accepted but it is not necessary that the turing machine will halt to indicate that the string is not accepted. in this case the user might end up in infinite wait by thinking that turing machine will halt string acceptance process is still in progress.

Complement of a language L means the complement form of given language L will contain all String except the strings that L contains.
Here L= {aa,bb}
It means complement of L can contain { λ, a, b, ab, ba, aaa, aab, aba, bba,..................}
Hence Option (D) is correct.

Context sensitive languages are closed under union, intersection, complement, concatenation, kleene star, ϵ-free Homomorphism, substitution( ϵ-free), inverse homomorphism, reverse, intersection with regular language.
Hence the correct option is (C)

{aⁿ bⁿ|n > 0} is a deterministic context free language but not regular because the given language requires a storing device and finite state machine does not contain any storing device. {aⁿ bⁿ|n > 0} can be accepted by a pushdown automata by pushing all the a’s on top of stack and when b’s come as input pop one “a” for each “b”.
The given language {aⁿ bⁿ a^{n. | n > 0} is a context sensitive language because for this language if we push a’s on top of stack and for every “b” we will pop a single “a” from top of stack and when after “b” again “a” will come as input then that time stack will be empty and we can’t compare number of a’s that come after “”b”. But we can have a linear bounded automata that can accept the given language by moving the read/write head accordingly.
Since the given language is CSL and CSL are closed under complement so given language can not be accepted by a deterministic pushdown automation {an bn an} is context sensitive language but not context free language, the logic is the same as above. SInce for given language can’t be accepted by pushdown automata, so it is not a context free language.}