Let A, B be algebras over algebraically closed field $\mathbb F$, and m be a maximal ideal in $A\otimes_\mathbb{F} B$ such that $A\otimes_\mathbb{F} B/m \cong \mathbb F$. Show that there are maximal ideals $a\subset A, b\subset B$ such that $m = a\otimes B + A\otimes b$.
Is this also true when m, a, b are prime ideals rather than maximal?
I presume that the isomorphism in $(A\otimes_\mathbb{F} B)/m \cong \mathbb F$ means an isomorphism of $\Bbb F$-algebras.
Your condition means that $m$ is the kernel of a $\Bbb F$-homomorphism $\phi:A\otimes B\to\Bbb F$. Such a homomorphism comes from a pair of $\Bbb F$-homomorphisms $\phi_1:A\to\Bbb F$ and $\phi_2:B\to\Bbb F$ and where $$\phi(x\otimes y)=\phi_1(x)\otimes\phi_2(y).$$ Let $a$ and $b$ be the kernels of $\phi_1$ and $\phi_2$.
For your second question, I presume you no longer want $(A\otimes_\mathbb{F} B)/m \cong \mathbb F$ since that necessitates the maximality of $m$. If so, consider the example $A=\Bbb F[X]$, $B=\Bbb F[Y]$ and $m=\left<X\otimes Y-1\right>$.