= Solution
{c}
Suppose, for a contradiction, that $P_2\cap\mathbb Z=(0)$. Then $P_1\cap\mathbb Z=(0)$ too. Both <prime ideals> are disjoint from the <multiplicatively closed set> $Y=\mathbb Z\setminus\{0\}$. The <localization> is
$$
Y^{-1}\mathbb Z[t]=\mathbb Q[t]:
$$
each fraction on the left has rational coefficients, and every <polynomial> with rational coefficients has a common nonzero integer denominator.
By <3/b>, the extended <prime ideals> form a strict inclusion
$$
\mathbb Q[t]P_1\subsetneq\mathbb Q[t]P_2.
$$
The first is nonzero, because $P_1$ contains a nonzero <polynomial> and the map $\mathbb Z[t]\to\mathbb Q[t]$ is <injective>.
The <polynomial ring> $\mathbb Q[t]$ is a <principal ideal domain>: in any nonzero <ideal>, a nonzero <polynomial> of least <polynomial degree> generates the <ideal>, since the <polynomial division algorithm> gives a remainder of smaller <polynomial degree>, which must be zero. By <3/a>, every nonzero <prime ideal> of this <principal ideal domain> is maximal. The strict inclusion of extended <prime ideals> is therefore impossible. We obtain
$$
\boxed{P_2\cap\mathbb Z\ne(0).}
$$
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