Which of the following statements is/are CORRECT?

Let us check the closure properties under intersection for each class of languages:

• (A) Regular languages: Regular languages are closed under intersection. If $L_1$ and $L_2$ are regular, their intersection $L_1 \cap L_2$ can be accepted by a product automaton, or by De Morgan's Law: $L_1 \cap L_2 = \overline{\overline{L_1} \cup \overline{L_2}}$. Thus, statement (A) is CORRECT.

• (B) Context-free languages (CFLs): CFLs are not closed under intersection. For example, $L_1 = \{a^n b^n c^m \mid n, m \ge 0\}$ and $L_2 = \{a^m b^n c^n \mid m, n \ge 0\}$ are both context-free, but their intersection $L_1 \cap L_2 = \{a^n b^n c^n \mid n \ge 0\}$ is not context-free. Thus, statement (B) is INCORRECT.

• (C) Recursive languages (Decidable): Recursive languages are closed under intersection. Given Turing machines (deciders) $M_1$ and $M_2$ for $L_1$ and $L_2$, a decider for $L_1 \cap L_2$ runs $M_1$ on input $w$; if $M_1$ accepts, it runs $M_2$ on $w$, accepting if and only if both accept. Since both deciders always halt, the composite machine also always halts. Thus, statement (C) is CORRECT.

• (D) Recursively enumerable languages (Turing-recognizable): Recursively enumerable languages are closed under intersection. Given Turing machines $M_1$ and $M_2$ recognizing $L_1$ and $L_2$, a Turing machine for $L_1 \cap L_2$ runs $M_1$ on input $w$, and if $M_1$ accepts, it then runs $M_2$ on $w$ (or runs them in parallel). It accepts if both accept. Thus, statement (D) is CORRECT.

Hence, the correct options are A, C, and D.