Language intersections (set theory) - Understanding better
In your last example, when trying to deduce $L \cap L'$, the trick is to trying to understand which strings in $L$ have even length. You can do this as follows:

$$L = \{a^{2n} b^m a^k | n,m &gt;= 0, k = m \pmod 2\}.$$

We can forgot about the first substring of $a$'s, because they always come in pairs (or there are none. In either case, they contribute an even number of characters. Now consider $\{b^m a^k | m &gt;= 0, k = m \pmod 2\}$. The length of the strings in this set is $m+k$. Since $k = m \pmod 2$, we have $m+k = m + m = 2m = 0 \mod 2$. Therefore, $m+k$ is always even, and so $b^m a^k$ always contributes an even number of characters as well. Since $even + even = even$, every string in $L$ is even. Therefore, $L \cap L' = L$.