For any two languages $L_1$ and $L_2$ such that $L_1$ is context-free and $L_2$ is recursively enumerable but not recursive, which of the following is/are necessarily true? I. $\bar{L}_1$ (complement of $L_1$) is recursive II. $\bar{L}_2$ (complement of $L_2$) is recursive III. $\bar{L}_1$ is context-free IV. $\bar{L}_1 \cup L_2$ is recursively enumerable

Let us analyze each statement based on language closure properties and the Chomsky hierarchy:

1. Statement I is TRUE: Every context-free language $L_1$ is a recursive (decidable) language. Since recursive languages are closed under complementation, $\bar{L}_1$ is also recursive.

2. Statement II is FALSE: $L_2$ is recursively enumerable (RE) but not recursive. If $\bar{L}_2$ were recursive, it would also be RE. By Post's theorem, if both a language and its complement are RE, then the language must be recursive, which contradicts the fact that $L_2$ is not recursive.

3. Statement III is FALSE: Context-free languages are not closed under complementation. Thus, $\bar{L}_1$ is not necessarily context-free.

4. Statement IV is TRUE: Since $\bar{L}_1$ is recursive, it is also recursively enumerable (RE). The class of RE languages is closed under union. Therefore, the union of two RE languages, $\bar{L}_1 \cup L_2$, is necessarily recursively enumerable.

Hence, statements I and IV only are necessarily true, which corresponds to option (D).