= Solution
In a <one-dimensional representation>, the scalar images of $Q$ and $P$ commute. The relation $PQP=Q^{-1}$ gives $R(Q)^2=1$, while $R(Q)^5=1$, so $R(Q)=1$ and $R(P)=\pm1$. These give exactly $R_0$ and $R_1$.
For $k=1,2$, define the two-dimensional <representations>
$$
D_k(Q)=\operatorname{diag}(\alpha^k,\alpha^{-k}),\qquad
D_k(P)=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
The same <invariant subspace> argument as in <1/a/solution> proves that each is <irreducible>. Set $R_2=D_1$ and $R_3=D_2$. Their <characters> differ at $Q$, so they are inequivalent. The <sum of squares of irreducible representation dimensions> is already $1^2+1^2+2^2+2^2=10$, proving completeness.
Put $s=2\cos(2\pi/5)=(\sqrt5-1)/2$ and $t=2\cos(4\pi/5)=-(\sqrt5+1)/2$. Taking <traces> yields the <character table>
$$
\boxed{\begin{array}{c|rrrr}
& I & Q,Q^4 & Q^2,Q^3 & Q^kP\\\hline
\text{class size}&1&2&2&5\\\hline
R_0&1&1&1&1\\
R_1&1&1&1&-1\\
R_2&2&s&t&0\\
R_3&2&t&s&0
\end{array}}
$$
The identities $s+t=-1$, $st=-1$, and $s^2+t^2=3$ also verify <character orthogonality> directly.
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