= Solution
Let $S=s=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Entrywise <complex conjugation> interchanges the diagonal entries of $r$ and fixes $s$, so
$$
SR_2(r)^*S^{-1}=R_2(r),\qquad SR_2(s)^*S^{-1}=R_2(s).
$$
Since $r,s$ generate the <group>, these identities hold for every element. Thus $S$ is an invertible <intertwiner>, proving
$$
\boxed{R_2^*\cong R_2.}
$$
The <character> of a <tensor product representation> is the product of the factor <characters>. Hence $\chi_{R_2\otimes R_2}=(4,1,0)$. Its <character inner products> with the three rows of <2/c/solution> give multiplicities $(1,1,1)$, so
$$
\boxed{R_2\otimes R_2\cong\mathbf1\oplus\varepsilon\oplus R_2.}
$$
For explicit <Clebsch-Gordan decomposition of group representations> <bases>, let $e_+,e_-$ be the coordinate <basis>. The <symmetric tensor> $e_+\otimes e_-+e_-\otimes e_+$ spans $\mathbf1$; the <antisymmetric tensor> $e_+\otimes e_--e_-\otimes e_+$ spans $\varepsilon$; and the ordered pair $e_-\otimes e_-,e_+\otimes e_+$ carries $R_2$, since $r$ acts by $\operatorname{diag}(\alpha,\alpha^2)$ and $s$ exchanges them. In particular $S^2R_2=\mathbf1\oplus R_2$ and $\Lambda^2R_2=\varepsilon$.
Multiplying by the <sign representation> preserves the <character> of $R_2$, because its <character> is zero on all three reflections. Thus $\varepsilon\otimes R_2\cong R_2$, and associativity gives
$$
\begin{aligned}
R_2^{\otimes3}
&\cong(\mathbf1\oplus\varepsilon\oplus R_2)\otimes R_2\\
&\cong R_2\oplus R_2\oplus(\mathbf1\oplus\varepsilon\oplus R_2).
\end{aligned}
$$
Therefore
$$
\boxed{R_2\otimes R_2\otimes R_2\cong\mathbf1\oplus\varepsilon\oplus3R_2.}
$$
The <dimensions> give $8=1+1+3\cdot2$. Its two one-dimensional summands can also be seen directly: $e_+^{\otimes3}+e_-^{\otimes3}$ and $e_+^{\otimes3}-e_-^{\otimes3}$ transform as $\mathbf1$ and $\varepsilon$, respectively.
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