We let $\varphi:A \rightarrow A'$ be a ring homomorphism and $\varphi^*:Spec\,A' \rightarrow Spec\,A$ the corresponding induced map between spectrum. Now we suppose that $Spec\,A'$ is irreducible and let $x$ be its (unique) generic point (i.e. $\overline{\{x\}}=Spec\,A'$). We want to prove that the closure $\overline{\varphi^*(Spec\,A')}$ is irreducible and $\varphi^*(x)$ is a generic point of $Spec\,A$ (i.e. $\overline{\varphi^*(x)}=Spec\,A$).
For the first part, the proof is not very hard. Note that $Spec\,A'$ is irreducible $\Leftrightarrow$ nilradical $rad(A')$ is a prime ideal. And from the property of the induced map between spectrum, we have $\overline{\varphi^*(Spec\,A')}=\overline{\varphi^*(V(rad(A'))}=V(\varphi^{-1}(rad(A')))$. We also know that $\overline{\varphi^*(Spec\,A')}$ is irreducible $\Leftrightarrow$ $I(V(\varphi^{-1}(rad(A'))))=\varphi^{-1}(rad(A'))$ is a prime ideal. From the definition of the induced map, we prove the first part. But I cannot prove the second part of the question. Since $Spec\,A'$ is irreducible, we know that the generic point $x=rad(A')$. So $\varphi^*(x)=\varphi^{-1}(rad(A'))$. If we want to prove that $\varphi^*(x)$ is a generic point $\Leftrightarrow$ $V(\varphi^{-1}(rad(A')))=Spec\,A=V(rad(A)).$ $\Leftrightarrow$ $\varphi^{-1}(rad(A')) \subseteq rad(A)$. But I wonder how to prove the last inclusion?
Your first claim about $\overline{\phi^{\ast}(\textrm{Spec}(A'))}$ being irreducible is a general fact from topology: a continuous image of an irreducible space is irreducible, and the closure of irreducible subset remains irreducible.
Your next claim is false without some further assumptions.