Let $L_1$ be a regular language and $L_2$ be a context-free language. Which of the following languages is/are context-free?

Let us analyze each option using set theory and closure properties of context-free languages (CFLs):

• Option (A): $L_1 \cap \overline{L_2}$ CFLs are not closed under complementation, so $\overline{L_2}$ may not be a CFL or even recursive. Thus, the intersection of a regular language and the complement of a CFL is not necessarily context-free.

• Option (B): $\overline{\overline{L_1} \cup \overline{L_2}}$ By De Morgan's laws:

  $$\overline{\overline{L_1} \cup \overline{L_2}} = \overline{\overline{L_1}} \cap \overline{\overline{L_2}} = L_1 \cap L_2$$

  The intersection of a regular language ($L_1$) and a context-free language ($L_2$) is always context-free. Thus, (B) is a CFL.

• Option (C): $L_1 \cup (L_2 \cup \overline{L_2})$ Since $L_2 \cup \overline{L_2} = \Sigma^*$:

  $$L_1 \cup (L_2 \cup \overline{L_2}) = L_1 \cup \Sigma^* = \Sigma^*$$

  $\Sigma^*$ is a regular language, and every regular language is also context-free. Thus, (C) is a CFL.

• Option (D): $(L_1 \cap L_2) \cup (\overline{L_1} \cap L_2)$ By distributive law of set theory:

  $$(L_1 \cap L_2) \cup (\overline{L_1} \cap L_2) = (L_1 \cup \overline{L_1}) \cap L_2 = \Sigma^* \cap L_2 = L_2$$

  Since $L_2$ is context-free, this language is context-free. Thus, (D) is a CFL.

Therefore, options B, C, and D are correct.