The rules are:
If A ≤ _p B
Rule 1: If B is recursive then A is recursive
Rule 2: If B is recursively enumerable then A is recursively enumerable
Rule 3: If A is not recursively enumerable then B is not recursively enumerable
Since X is recursive and recursive language is closed under compliment, so is also recursive.
: By rule 3, W is not recursively enumerable.
: By rule 1, Z is recursive.
Hence, W is not recursively enumerable and Z is recursive.

L2 − L1 means L2 ∩ L1^c , since L1 is recursive hence L1^c must also be recursive, So L2∩L1^c is equivalent to (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2− L1 must be recursive enumerable. Hence A is always true.
L1 − L3 means L1 ∩ L3^c , since recursive enumerable is not closed under complement, so L3^c may or may not be recursive enumerable, hence we cannot say that L1 − L3 will always be recursive enumerable. So B is not necessarily true always.
L2 ∩ L1 means (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2∩ L1 must be recursive enumerable. Hence C is always true.
L2 ∪ L1 means (Recursive enum ∪ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∪ Recursive enum) and recursive enum is closed under union, so L2 ∪ L1 must be recursive enumerable. Hence D is always true.

If L is recursive enumerable, then it implies that there exist a Turing Machine (lets say M1) which always HALT for the strings which is in L.
If are recursively enumerable, then it implies that there exist a Turing Machine (lets say M2) which always HALT for the strings which is NOT in L(as L is complement of Since we can combine both Turing machines (M1 and M2) and obtain a new Turing Machine (say M3) which always HALT for the strings if it is in L and also if it is not in L. This implies that L must be recursive.

Every NFA can be converted into DFA (as there exist a standard procedure to convert NFA into DFA). Also, every non-deterministic Turing machine can be converted to an equivalent deterministic Turing machine. Every regular language is also a CFL , since if a language can be recognized by Finite automata, then it must also be recognize by PDA (as PDA is more powerful than FA). But every subset of recursively enumerable need not be recursive.

Recursive languages are closed under complementation.
But, recursive enumerable languages are not closed under complementation.
If L₁ is recursive, then L₁', is also recursive.
If L₂ is recursive enumerable, then L₂', is not recursive enumerable language.

Given that
L₁ = recursively enumerable language
L₂ over Σ∪{#} as {w_i#w_j | w_i, w_j ∈ L₁, i < j}
S₁ is True.
If L₁ is recursive then L₂ is also recursive. Because, to check w = w_i#w_j belongs to L₂, and we give w_i and w_j to the corresponding decider L₁, if both are to be accepted, then w∈L₁ and not otherwise.
S₂ is also True:
With the corresponding decider L₂ we need to make decide L₁.

Here, we have two possibilities, whether
P = NP (or) P != NP
→ If P=NP then L=(0+1)* which is recular, then it is recursive.
→ If P!=NP then L becomes ɸ which is also regular, then it is recursive.
So, finally L is recursive.