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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, $L' \subseteq L$
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