Say, $A$ is universal set , $A=\Sigma^* =(a+b)^*$
It is clear that $L1$ is subset of $A$, and $L2$ is subset of $L1$.
it is clear from the diagram that $L1 \cup\overline{L2} = L =\Sigma^{*} =(a+b)^{*}$
Note :
It is very often to see, that we thought $\overline{L2}$ = $\overline{\{ a^mb^n |m =n\}}$= {$a^mb^n |m\neq n$} but remember $\overline{L2}$ also contain simple $a$, or $a^*$ or $b^*$ or $b^*a^*$ or strings as $aba$ or many things.
Actually $\overline{L2}=\Sigma^* -\{a^nb^n\}$