$a$ is true because every finite language is regular language.
$b$ is true for every regular expression $r,s.$
Let $R, S$ be any two regular expressions, then :
$$
\begin{aligned}
&(R+S)^{\ast }=\left(R^{\ast }+S^{\ast }\right)^{\ast }=\left(R^{\ast } S^{\ast }\right)^{\ast }=\left(R^{\ast } S\right)^{\ast } R^{\ast }=R^{\ast }\left(S R^{\ast }\right)^{\ast } . \\
&R(S R)^{\ast }=(R S)^{\ast } R .
\end{aligned}
$$
