Difference between $a^{2n} b^n$ and $n_a(w) = 2n_b(w)$
The usual definition is that $n_a(w)$ refers to the number of $a$s in $w$.

Based on this definition, $L_1$ consists exactly of words that begin with an even number of $a$s followed by half as many $b$s (no $c$s are permitted). Therefore, you have $\varepsilon, aab, aaaabb, aaaaaabbb,... \in L_1$.

Whereas $L_2$ consists of words containing an even number of $a$s and half as many $b$s (in any order and with an arbitrary number of $c$ in between).

Thus every word in $L_1$ is also in $L_2$ (formally $L_1 \subseteq L_2$), but there are words in $L_2$ that are not in $L_1$:


For some $n \geq 0$, consider words of the form $b^n a^{2n}$. They are in $L_2$, because they contain an even number of $a$s and half as many $b$s, but if $n &gt; 0$, then it does not match $a^{2n} b^n$, because the $a$s come after the $b$s. You have $\varepsilon, baa,bbaaa,bbbaaaaaa,... \in L_2$, but $baa,bbaaa,bbbaaaaaa \notin L_1$.
The same argument applies for $w = (aab)^n$. You have $\varepsilon, aab,aabaab,aabaabaab \in L_2$, but $aabaab,aabaabaab \notin L_1$.
A word can have an even number of $a$s and half as many $b$s and also contain $c$s. For example, $a^{2n} b^n c \in L_2$. But the words in $L_1$ by definition only contain $a$s and $b$s.