Let us examine each statement carefully using standard results from automata and computability theory.
(A) False.
A recursively enumerable (r.e.) language need not have a recursively enumerable complement. Since recursive (decidable) languages are exactly those whose language and complement are both r.e., it follows that not every subset of an r.e. language is recursive. Hence, the statement is false.
(B) True.
If a language \( L \) and its complement \( \overline{L} \) are both recursively enumerable, then there exist Turing machines that semi-decide both. This implies that membership in \( L \) can be decided by running both machines in parallel and accepting whichever halts first. Therefore, \( L \) is recursive. Hence, this statement is true.
(C) True.
Although context-free languages are not closed under complementation, every context-free language is decidable (recursive). Since recursive languages are closed under complementation, the complement of a context-free language must be recursive, even though it may not be context-free. Thus, the statement is true.
(D) True.
Regular languages are closed under intersection. Moreover, every regular language is also deterministic context-free (DCFL). Since the intersection of two regular languages is regular, it is also deterministic context-free. Therefore, this statement is true.
Final Answer:
The correct statements are (B), (C), and (D).