 Solution

The inverse image $I={\theta }^{-1}(J)$ is an ideal, by preservation of differences and multiplication. It is proper since $\theta (1)=1\notin J$. If $ab\in I$, then $\theta (a)\theta (b)\in J$; primality of $J$ gives $\theta (a)\in J$ or $\theta (b)\in J$, hence $a\in I$ or $b\in I$. Therefore the inverse image is prime.

For maximality, take the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ and $J=(0)$. This is a maximal ideal of the field $\mathbb{Q}$, but its inverse image is $(0)$ in $\mathbb{Z}$, strictly contained in the proper ideal $2\mathbb{Z}$, so it is not maximal. Preservation of identity matters for proper inverse images: without it the zero homomorphism could pull a proper ideal back to the whole ring.