Solution

Given language:

$L={a^n b^n\mid n\ge 0}$

This is a context free language.

Now check each statement.

For $S_1$:

$L^2=L\cdot L$

Context free languages are closed under concatenation.

So, $L^2$ is also context free.

Hence, $S_1$ is true.

For $S_2$:

$L^k$ means concatenation of $L$ with itself $k$ times.

For any fixed $k\ge 1$, context free languages are closed under finite concatenation.

So, $L^k$ is context free.

Hence, $S_2$ is true.

For $S_3$:

$L^*$ is context free because context free languages are closed under Kleene star.

Also, for this specific language $L={a^n b^n\mid n\ge 0}$, its complement $\overline{L}$ is also context free.

So, $\overline{L}$ and $L^*$ both are context free languages.

Hence, $S_3$ is true.

Therefore, all three statements $S_1,\ S_2$ and $S_3$ are correct.