S1 is true. $L$ is reduced to a decidable problem and hence $L$ also becomes decidable.
S2 is false. Enumerator for $L$ also enumerates strings in $L'$, but it might also enumerate strings not in $L'$. For example, take $L$ as $\Sigma^*$ and take $L'$ as a non recursively enumerable language over $\Sigma$. Now, $L' \subseteq L$