Is $(a+b)^*$ closed under reversal?
If an arbitrary language $L$ is closed under reversal ($L=L^R$), then $L^*$ is also closed under reversal, since

$$
\begin{array}{ll}
  L^R &amp;= \{ w_1 \ldots w_n \ |\ \forall i.\ w_i \in L \}^R \\
      &amp;= \{ (w_1 \ldots w_n)^R \ |\ \forall i.\ w_i \in L \} \\
      &amp;= \{ w_n^R \ldots w_1^R \ |\ \forall i.\ w_i \in L \} \\
      &amp;= \{ x_1 \ldots x_n \ |\ \forall i.\ x_i^R \in L \} \\
      &amp;= \{ x_1 \ldots x_n \ |\ \forall i.\ x_i \in L^R \} \\
      &amp;= \{ x_1 \ldots x_n \ |\ \forall i.\ x_i \in L \} \\
\end{array}
$$

So, being $\{a,b\}$ trivially closed under reversal (provided $a,b$ are letters, not words), $\{a,b\}^*$ is such.