Yes. The set of symmetric matrices $Q$ such that $AQ=QB$ is a vector space $V$ over $F$. We know that $V\otimes _FK\neq 0$, hence $V\neq 0$. The invertible matrices form a Zariski open subset $U$ of $V$ (defined by $\det\neq 0$). Again we know that $U\otimes _FK$ is a nonempty Zariski open subset of $V\otimes _FK$; since $V$ is Zariski dense in $V\otimes _FK$, $U$ is nonempty.

Yes, if $F$ is infinite. The set of symmetric matrices $Q$ such that $AQ=QB$ is a vector space $V$ over $F$. We know that $V\otimes _FK\neq 0$, hence $V\neq 0$. The polynomial $\det_{|F}$ is not identically zero in $V\otimes _FK$, hence in $V$ since $F$ is infinite. Thus there exists $Q\in V$ invertible.

Yes. The set of symmetric matrices $Q$ such that $AQ=QB$ is a vector space $V$ over $F$. We know that $V\otimes _FK\neq 0$, hence $V\neq 0$.

Yes. The set of symmetric matrices $Q$ such that $AQ=QB$ is a vector space $V$ over $F$. We know that $V\otimes _FK\neq 0$, hence $V\neq 0$. The invertible matrices form a Zariski open subset $U$ of $V$ (defined by $\det\neq 0$). Again we know that $U\otimes _FK$ is a nonempty Zariski open subset of $V\otimes _FK$; since $V$ is Zariski dense in $V\otimes _FK$, $U$ is nonempty.