Proving iff statement with reductions
There is a more general theorem which says that if $A$ is a decidable language and $B$ is a decidable language that is not trivial (in the sense that $B\ne\Sigma^*$ and $B\ne\emptyset\, $), then $A\le_m B$. Roughly speaking, this says that all non-trivial decidable languages are many-to-one reducible to each other, meaning that $\le_m$ is too coarse a metric to distinguish between decidable languges.

In your problem, suppose that $L\subseteq \Gamma^*$ and define a mapping $f:\Gamma^*\rightarrow\Sigma^*$ defined by:
$$
f(x) =
 \begin{cases}
   \mathtt{aa} &amp; \text{if $x\in L$}\\
   \mathtt{a} &amp; \text{if $x\notin L$}
 \end{cases}
$$
for some $\mathtt{a}\in\Sigma^*$. This is a total computable function (since by definition, we can use the computable decider program to determine $x$'s membership in $L\, $) and it's easy to verify that $x\in L$ iff $f(x)\in A$, which is all you need to show $L\le_m A$. As Yuval noted, you don't need any results about regular or context-free languages to answer this question.