$L_1=\{a^nba^n\;|\; n>0\}$ is CFL
push $a's$ in stack ,do nothing with $b$ , then pop $a's$ from stack on reading $a's$
$L_2=\{a^nba^nb^{n+1}\;|\; n>0\}$ is CSL.
Push $a's$ in stack, donothing on $b$ , pop $a's$ from stack on reading $a's$, then we left with $b's$ with empty stack, then we can't ensure these $b's$ are one more that earlier $a's$.
Option $A$, $L_1$ is context free and $L_2$ is not context free language.