= Solution
Since $p$ and $q$ are coprime, <Bézout's identity for polynomials> gives $a,b\in F[x]$ with $ap+bq=1$. For every $v\in V$,
$$
v=b(T)q(T)v+a(T)p(T)v.
$$
The first summand belongs to $U$: applying $p(T)$ gives $b(T)p(T)q(T)v=0$. The second belongs to $W$ by the same calculation. Here all <polynomials> in $T$ commute, and $p(T)q(T)=f(T)=0$. Thus $V=U+W$.
If $v\in U\cap W$, both terms on the right of the identity vanish, so $v=0$. Therefore
$$
\boxed{V=U\oplus W.}
$$
Finally, $p(T)Tv=Tp(T)v=0$ whenever $v\in U$; hence $T(U)\subseteq U$. Replacing $p$ by $q$ proves $T(W)\subseteq W$. These are <invariant subspaces>; this is the coprime-factor case of the <primary decomposition theorem for linear maps>.
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