Which of the following languages are context-free? $L_1 = \{a^m b^n a^n b^m \mid m, n \ge 1\}$ $L_2 = \{a^m b^n a^m b^n \mid m, n \ge 1\}$ $L_3 = \{a^m b^n \mid m = 2n + 1\}$

1. Language $L_1 = \{a^m b^n a^n b^m \mid m, n \ge 1\}$: The dependencies in $L_1$ are nested in a LIFO manner (the inner $b^n a^n$ can be matched first, and the outer $a^m b^m$ is matched afterward). A pushdown automaton (PDA) can push all initial $a$'s onto the stack, then push all $b$'s, pop a $b$ for each subsequent $a$ to match $b^n a^n$, and finally pop an $a$ for each trailing $b$ to match $a^m b^m$. Because a single stack suffices, $L_1$ is context-free.

2. Language $L_2 = \{a^m b^n a^m b^n \mid m, n \ge 1\}$: The dependencies in $L_2$ are cross-serial (the first block of $a$'s matches the second block of $a$'s, and the first block of $b$'s matches the second block of $b$'s). A single stack cannot maintain and verify both counts simultaneously. By the Pumping Lemma for Context-Free Languages, $L_2$ is not context-free.

3. Language $L_3 = \{a^m b^n \mid m = 2n + 1\}$: $L_3 = \{a^{2n+1} b^n \mid n \ge 0\}$. A PDA can easily match one $b$ for every two $a$'s popped from the stack (ignoring the single extra leading $a$). Thus, $L_3$ is context-free (generated by $S \to aS^{\prime}b$, $S^{\prime} \to aaS^{\prime}b \mid a$).

Therefore, only $L_1$ and $L_3$ are context-free, which matches option (B).