We will use that smooth ring maps are flat (Algebra, Lemma \ref{algebra-lemma-smooth-syntomic}), that localizations are flat (Algebra, Lemma \ref{algebra-lemma-flat-localization}), that compositions of flat ring maps are flat (Algebra, Lemma \ref{algebra-lemma-composition-flat}), that base change of a flat ring map is flat (Algebra, Lemma \ref{algebra-lemma-flat-base-change}), that flat local homomorphisms are faithfully flat (Algebra, Lemma \ref{algebra-lemma-local-flat-ff}), that (strict) henselization is flat (Lemma \ref{lemma-dumb-properties-henselization}), and Going down for flat ring maps (Algebra, Lemma \ref{algebra-lemma-flat-going-down}). \medskip\noindent Proof of (2). Let $A^h$, $B^h$ be the henselizations of $A$, $B$. Then $B^h$ is the henselization of $A^h \otimes_A B$ at the unique maximal ideal lying over $\mathfrak m_B$, see Algebra, Lemma \ref{algebra-lemma-henselian-functorial-improve}. Thus we may and do assume $A$ is henselian. Since $A \to B \to B^h$ is flat, every minimal prime of $B^h$ lies over a minimal prime of $A$ and since $A \to B^h$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^h$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^h$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; the $A$ is still henselian by Algebra, Lemma \ref{algebra-lemma-quotient-henselization}. By Lemma \ref{lemma-unibranch} we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain. Of course $A'$ is a normal domain. By Algebra, Lemma \ref{algebra-lemma-normal-goes-up} we see that $A' \otimes_A B^h$ is a normal ring (the lemma just gives it for $A' \otimes_A B$, to go up to $A' \otimes_A B^h$ use that $B^h$ is a colimit of \'etale $B$-algebras and use Algebra, Lemma \ref{algebra-lemma-colimit-normal-ring}). By Lemma \ref{lemma-local-tensor-with-integral} we see that $A' \otimes_A B^h$ is local (this is where we use the assumption on the residue fields of $A$ and $B$). Hence $A' \otimes_A B^h$ is a local normal ring, hence a local domain. Since $B^h \subset A' \otimes_A B^h$ by flatness of $A \to B^h$ we conclude that $B^h$ is a domain as desired. \medskip\noindent Proof of (1). Let $A^{sh}$, $B^{sh}$ be strict henselizations of $A$, $B$. Then $B^{sh}$ is a strict henselization of $A^h \otimes_A B$ at a maximal ideal lying over $\mathfrak m_B$ and $\mathfrak m_{A^h}$, see Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial-improve}. Thus we may and do assume $A$ is strictly henselian. Since $A \to B \to B^{sh}$ is flat, every minimal prime of $B^{sh}$ lies over a minimal prime of $A$ and since $A \to B^{sh}$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^{sh}$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^{sh}$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; then $A$ is still strictly henselian by Algebra, Lemma \ref{algebra-lemma-quotient-strict-henselization}. By Lemma \ref{lemma-geometrically-unibranch} we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain whose residue field is a purely inseparable extension of the residue field of $A$. Of course $A'$ is a normal domain. By Algebra, Lemma \ref{algebra-lemma-normal-goes-up} we see that $A' \otimes_A B^{sh}$ is a normal ring (the lemma just gives it for $A' \otimes_A B$, to go up to $A' \otimes_A B^{sh}$ use that $B^{sh}$ is a colimit of \'etale $B$-algebras and use Algebra, Lemma \ref{algebra-lemma-colimit-normal-ring}). By Lemma \ref{lemma-local-tensor-with-integral} we see that $A' \otimes_A B^{sh}$ is local (since $A \subset A'$ induces a purely inseparable residue field extension). Hence $A' \otimes_A B^{sh}$ is a local normal ring, hence a local domain. Since $B^{sh} \subset A' \otimes_A B^{sh}$ by flatness of $A \to B^{sh}$ we conclude that $B^{sh}$ is a domain as desired.