= Solution
Let $T(v\otimes w)=w\otimes v$ on the <tensor product> $V\otimes V$. The <linear map> $R(g)\otimes R(g)$ commutes with $T$, so preserves its $+1$ and $-1$ <eigenspaces>, namely the <symmetric square> $S^2V$ and the <exterior square> $\wedge^2V$. Restricting the <tensor product representation> to these <invariant subspaces> defines $S^2R$ and $\wedge^2R$. The <representation> identity follows by restricting $(R(g)\otimes R(g))(R(h)\otimes R(h))=R(gh)\otimes R(gh)$.
For a <finite group>, $g$ has finite order $m$, so $R(g)^m=I$. Over the <complex numbers>, the <polynomial> $t^m-1$ has distinct roots; hence $R(g)$ is <diagonalizable>. Choose an <eigenbasis> $v_1,\ldots,v_n$ with <eigenvalues> $\lambda_1,\ldots,\lambda_n$. A <basis> of the <symmetric square> consists of $v_i\otimes v_i$ and $v_i\otimes v_j+v_j\otimes v_i$ for $i<j$. A <basis> of the <exterior square> consists of $v_i\otimes v_j-v_j\otimes v_i$ for $i<j$. Their <eigenvalues> are respectively $\lambda_i^2,\lambda_i\lambda_j$ and $\lambda_i\lambda_j$. Taking <traces>,
$$
\chi_{S^2R}(g)=\sum_i\lambda_i^2+\sum_{i<j}\lambda_i\lambda_j,
\qquad \chi_{\wedge^2R}(g)=\sum_{i<j}\lambda_i\lambda_j.
$$
But $\chi(g)^2=\sum_i\lambda_i^2+2\sum_{i<j}\lambda_i\lambda_j$ and $\chi(g^2)=\sum_i\lambda_i^2$. Thus the <characters> are
$$
\boxed{\chi_{S^2R}(g)=\tfrac12[\chi(g)^2+\chi(g^2)],\qquad
\chi_{\wedge^2R}(g)=\tfrac12[\chi(g)^2-\chi(g^2)].}
$$
At $g=e$, these also give the <dimensions> $n(n+1)/2$ and $n(n-1)/2$.
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