Kleene star operations
Hint:

You are right on (a). The proof involves key results for Kleene star as follows.
$$
\Sigma^{*}=\bigcup_{n\geqslant1}\Sigma^{n}\quad\text{and}\quad L\Sigma^{*}=\bigcup_{n\geqslant1}L\Sigma^{n}\quad\text{and}\quad \Sigma^{*}L=\bigcup_{n\geqslant1}\Sigma^{n}L
$$
Where $\Sigma^{n}$ can be considered as the set of the concatenation of i strings in $\Sigma$, and $L\subset \Sigma^{*}$.

For (a), $L_1\Sigma^{*}\subset\Sigma^{*}$ since for all $n,L_1\Sigma^{n}\subset\Sigma^{n}$. If $\Lambda\in L_1$, then $n,L_1\Sigma^{n}=\Sigma^{n}$ for all $n$. Thus $L_1\Sigma^{*}=\Sigma^{*}$.

For (b), notice that if $\Lambda\in L_1$ and $L_2$, then $L_1\Sigma^{*}=\Sigma^{*}$ and $\Sigma^{*}L_2=\Sigma^{*}$. Thus
$L_1\Sigma^{*}L_2=\Sigma^{*}L_2=\Sigma^{*}$.