If $A \in \mathrm{RE}$ and $A \leq_m \overline{A}$ then $A\in \mathrm{R}$
It follows immediately from the definition of a mapping reduction that a reduction $f$ from $A$ to $B$ is also a reduction from $\overline{A}$ to $\overline{B}$. Indeed, if $f$ is a reduction from $A$ to $B$, then, in particular, $x\in A$ iff $f(x)\in B$, which is equivalent to $x\notin A$ iff $f(x)\notin B$, which is equivalent to $x\in \overline{A}$ iff $f(x) \in \overline{B}$. (I only used the fact that $x \to y$ is equivalent to $\neg y \to \neg x$).

This follows form the reduction theorem: if $A \leq _m B$, then $B \in RE \to A\in RE$. Try to prove it. Hint: use a machine that computes the reduction, and a machine that recognizes $B$ in order to define a machine that recognizes $A$.

To deduce that $A \in R$, they've used the fact that $R \supseteq RE\cap coRE$ (In fact, equality holds). Hint:  If $M_{A}$ and $M_{\overline{A}}$ are machines that recognize $A$ and $\overline{A}$, respectively, then you know that for every word $w$, at least one of $M_{A}$ and $M_{\overline{A}}$ halts on $w$.