L1 will not be CFL also as it needs more than one stack to count. L1/L2 gives concatenation and the result will be $a^{n} b^{2n} c^{4n}$ which is again non regular. Am I right? I am little bit confused for L1/L2 case as for some y, xy belongs to L1.
Help appreciated :)
You are correct in thinking that $L_1$ is not even context-free, though the reason that you give is not a proof; it’s not too hard to find a proof using the pumping lemma for context-free languages, however, and it’s even easier to prove that $L_1$ is not regular using the pumping lemma for regular languages.
It’s true that $L_1\subseteq L_1/L_2$, for the reason that you gave in your second comment: $\epsilon\in L_2$, so for each $x\in L_1$ we can set $y=\epsilon\in L_2$ and have $xy\in L_1$, which implies that $x\in L_1/L_2$. That’s not enough to show that $L_1/L_2$ is not regular, however: after all, $L_1\subseteq\Sigma^*$, and $\Sigma^*$ is certainly regular. To decide whether $L_1/L_2$ is regular, we must see what else is in $L_1/L_2$.
Suppose that $xb^nc^{2n}\in L_1$. If $n=0$, then $x\in L_1$, and if $n>0$, then $x$ must be $a^n$. This shows that
$$L_1/L_2=L_1\cup\{a^n:n\ge 0\}\;.$$
To finish the problem, you can use the pumping lemma for regular languages to show that this is not regular.