Complement and Context Free Surprising
If two languages $A$, $B$ are complements of each other, then their disjoint union $A \uplus B$ must be the set of all strings $\Sigma^{\ast}$ over the given alphabet $\Sigma$. That is, every string in $\Sigma^{\ast}$ has to be in $A$ or $B$ but not both. So if we want to show that $L_{1} \neq \overline{L_{2}}$, all we have to do is find a string that is either in both or neither.

In this case it's fairly trivial. All strings in $L_{1}$ and $L_{2}$ have a $c$ in them, so pick any string that doesn't have a $c$ it in. For example $\varepsilon$, it's in neither $L_{1}$ nor $L_{2}$.

A construction for a PDA for $L_{1}$ is given here.

As for proving $L_{2}$ is not context free, everything you need is here.