= Solution
The <symmetric power> $S^nV$ is the quotient of $V^{\otimes n}$ identifying tensors which differ by permutation of the factors. The <exterior power> $\Lambda^nV$ is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a <basis> $e_1,\ldots,e_d$, bases consist respectively of monomials $e_1^{a_1}\cdots e_d^{a_d}$ with $\sum a_i=n$, and wedge products $e_{i_1}\wedge\cdots\wedge e_{i_n}$ with $i_1<\cdots<i_n$. Thus
$$
\boxed{\dim S^nV=\binom{d+n-1}{n},\qquad\dim\Lambda^nV=\binom dn,}
$$
with the second dimension zero for $n>d$.
A finite-order group element is diagonalizable over $\mathbb C$, because its minimal <polynomial> divides $X^m-1$, which has no repeated roots. Choose an <eigenbasis> with <eigenvalues> $\lambda_1,\ldots,\lambda_d$. The induced <eigenvalues>, including multiplicities, are
$$
\boxed{\lambda_1^{a_1}\cdots\lambda_d^{a_d}\ (a_i\geq0,\ \sum a_i=n)\text{ on }S^nV;\quad\lambda_{i_1}\cdots\lambda_{i_n}\ (i_1<\cdots<i_n)\text{ on }\Lambda^nV.}
$$
The zeroth powers are the one-dimensional trivial representation, with the empty product equal to $1$.
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