mapping reductions from R to RE
If $L_2 \neq \Sigma^*$ and $L_2 \neq \emptyset$ then $L_1 \in R$ and $L_2 \in RE$ implies $L_1 \le_m L_2$.
Let $T$ be a Turing machine that decides $L_1$. Let $a,b \in \Sigma^*$ such that $a \in L_2$ and $b \not\in L_2$.
For $x \in \Sigma^*$, define $\phi(x) =
\begin{cases}
a &amp; \text{ if $T(x)$ accepts }\\
b &amp; \text{ if $T(x)$ rejects }
\end{cases}$.
It is easy to check that $\phi$ is a mapping reduction from $L_1$ to $L_2$.
If $L_2 = \Sigma^*$ then $L_1 \in R$ and $L_2 \in RE$ does not imply $L_1 \le_m L_2$.
This can be seen, e.g., by choosing $L_1 = \emptyset$.
If $L_2 = \emptyset$ then $L_1 \in R$ and $L_2 \in RE$ does not imply $L_1 \le_m L_2$.
This can be seen, e.g., by choosing $L_1 = \Sigma^*$.