= Solution
Write $m_1,m_2$ for the <minimal polynomials> of the two <restrictions>. Since $p(T)$ vanishes on $U$ and $q(T)$ vanishes on $W$, part (b) gives
$$
m_1\mid p,\qquad m_2\mid q.
$$
Hence $m_1,m_2$ are coprime. By the direct-sum decomposition, a <polynomial> annihilates $T$ exactly when it annihilates both <restrictions>, which happens exactly when it is divisible by both $m_1$ and $m_2$. Consequently
$$
m_T=\operatorname{lcm}(m_1,m_2)=m_1m_2.
$$
If $m_1=p$ and $m_2=q$, this immediately gives $m_T=pq=f$. Conversely, if $m_T=pq$, the <monic polynomials> $p/m_1$ and $q/m_2$ have product one, so each is one. Thus
$$
\boxed{m_T=pq\iff m_{T|_U}=p\text{ and }m_{T|_W}=q.}
$$
This also handles a zero summand, whose <minimal polynomial> is one; a nonconstant factor cannot be its <minimal polynomial>.
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