Let $I \subset A$ be the ideal of nilpotents. We have $(A/I)^h = A^h/IA^h$ by (Algebra, Lemma \ref{algebra-lemma-quotient-henselization}). The spectra of $A$, $A^h$, $A'$, and $(A')^h$ are the same as the spectra of $A/I$, $A^h/IA^h$, $A'$, and $(A')^h = A' \otimes_{A/I} A^h/IA^h$. Thus we may replace $A$ by $A_{red} = A/I$ and assume $A$ is reduced. Then $A \subset A'$ which we will use below without further mention. \medskip\noindent Proof of (1). As $A'$ is integral over $A$ we see that $(A')^h$ is integral over $A^h$. By going up (Algebra, Lemma \ref{algebra-lemma-integral-going-up}) every maximal ideal of $A'$, resp.\ $(A')^h$ lies over the maximal ideal $\mathfrak m$, resp.\ $\mathfrak m^h$ of $A$, resp.\ $A^h$. Thus (1) follows from the isomorphism $$ (A')^h \otimes_{A^h} \kappa^h = A' \otimes_A A^h \otimes_{A^h} \kappa^h = A' \otimes_A \kappa $$ because the residue field extension $\kappa \subset \kappa^h$ induced by $A \to A^h$ is trivial. We will use below that the displayed ring is integral over a field hence spectrum of this ring is a profinite space, see Algebra, Lemmas \ref{algebra-lemma-integral-over-field} and \ref{algebra-lemma-ring-with-only-minimal-primes}. \medskip\noindent Proof of (3). The ring $A'$ is a normal ring and in fact a finite product of normal domains, see Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}. Since $A^h$ is a filtered colimit of \'etale $A$-algebras, $(A')^h$ is filtered colimit of \'etale $A'$-algebras hence $(A')^h$ is a normal ring by Algebra, Lemmas \ref{algebra-lemma-normal-goes-up} and \ref{algebra-lemma-colimit-normal-ring}. Thus every local ring of $(A')^h$ is a normal domain and we see that every maximal ideal contains a unique minimal prime. By Lemma \ref{lemma-integral-over-henselian-pair} applied to $A^h \to (A')^h$ we see that $((A')^h, \mathfrak m(A')^h)$ is a henselian pair. If $\mathfrak q \subset (A')^h$ is a minimal prime (or any prime), then the intersection of $V(\mathfrak q)$ with $V(\mathfrak m (A')^h)$ is connected by Lemma \ref{lemma-irreducible-henselian-pair-connected} Since $V(\mathfrak m (A')^h) = \Spec((A')^h \otimes \kappa^h)$ is a profinite space by we see there is a unique maximal ideal containing $\mathfrak q$. \medskip\noindent Proof of (2). The minimal primes of $A'$ are exactly the primes lying over a minimal prime of $A$ (by construction). Since $A' \to (A')^h$ is flat by going down (Algebra, Lemma \ref{algebra-lemma-flat-going-down}) every minimal prime of $(A')^h$ lies over a minimal prime of $A'$. Conversely, any prime of $(A')^h$ lying over a minimal prime of $A'$ is minimal because $(A')^h$ is a filtered colimit of \'etale hence quasi-finite algebras over $A'$ (small detail omitted). We conclude that the minimal primes of $(A')^h$ are exactly the primes which lie over a minimal prime of $A$. Similarly, the minimal primes of $A^h$ are exactly the primes lying over minimal primes of $A$. By construction we have $A' \otimes_A Q(A) = Q(A)$ where $Q(A)$ is the total fraction ring of our reduced local ring $A$. Of course $Q(A)$ is the finite product of residue fields of the minimal primes of $A$. It follows that $$ (A')^h \otimes_A Q(A) = A^h \otimes_A A' \otimes_A Q(A) = A^h \otimes_A Q(A) $$ Our discussion above shows the spectrum of the ring on the left is the set of minimal primes of $(A')^h$ and the spectrum of the ring on the right is the is the set of minimal primes of $A^h$. This finishes the proof.