Intersection of two particular languages
$$L_1 = \{a^{n+1}b^{2n}a^{n+1} \mid n \ge 0\}$$

$$L_2 = \{w \mid (\#_a(w) + \#_b(w)) \bmod 2 = 1\}$$

I assume that $\#_a(w)$ means the number of $a$s in $w$.

First, I would write out some $w$ in $L_1$ and $L_2$:

$$L_1 = \{aa, aabbaa, aaabbbbaaa, \ldots\}$$

$$L_2 = \{a, b, aaa, aab, aba, abb, baa, bab, bba, bbb, aaaaa, \ldots\}$$

From this, we see that words in $L_1$ are always even length ($(n+1) + (2n) + (n+1) = 4n + 2 = 2(2n+1)$ and odd + even + odd = even) and words in $L_2$ are always odd length ($(a+b)\bmod2 = 1$).

This means $L_1 \cap L_2 = \emptyset$.



Before OP's edit to post ($L_2 = \{w \mid (\#_a(w) + \#_b(w)) \bmod 2 = 0\}$):

First, I would write out some $w$ in $L_1$ and $L_2$:

$$L_1 = \{aa, aabbaa, aaabbbbaaa, \ldots\}$$

$$L_2 = \{aa, ab, ba, bb, aaaa, aaab, aaba, \ldots\}$$

From this, we see that words in $L_1$ are always even length ($(n+1) + (2n) + (n+1) = 4n + 2 = 2(2n+1)$ and odd + even + odd = even) and words in $L_2$ are always even length ($(a+b)\bmod2 = 0$).

This means $L_1 \cap L_2 = L_1$ (because $L_1 \subset L_2$).