Tensor product of $k$-algebras such that $\bigcap_{\frak{m}}\mathfrak m=(0)$ where $\frak{m}$ satisfy $A/{\frak m}\cong k$ Sorry for my bad English.
Let $k$ be a field, and $A$ be a $k$-algebra (if necessary, of finite type).
Consider the following condition concerned with $A$.

(Condition) $A$ satisfies $\bigcap_{\frak m}{\frak m}=(0)\subset  A$, where $\frak{m}$ runs over all maximal ideals such that $A/{\frak m}\cong k$.

If $A$ satisfies the above condition, then $A$ is reduced.
First question: Is the above condition named or famous?
Second question: Does $A\otimes_k B$ satisfy that condition if $A, B$ satisfy?
 A: I don't know about the first question, but here's a solution to the second. Note that a $k$-algebra $C$ satisfies your condition if and only if, for any $c\in C$, if $\theta(c)=0$ for every $k$-algebra morphism $C\to k$ then $c=0$.
Okay, so suppose $A$ and $B$ satisfy the condition, let $C=A\otimes_k B$, and suppose $\theta(c)=0$ for every $k$-algebra morphism $C\to k$; we wish to show that $c=0$. Without loss of generality we may write $c=\sum_{i\in[n]}\lambda_{i} (a_i\otimes b_i)$ where $\{b_1,\dots,b_n\}$ is linearly independent over $k$ and all $\lambda_{i}\in k^\times$. I claim $a_i=0$ for all $i\in[n]$, which will show the desired result; by the hypothesis on $A$, it suffices to show that $\varphi(a_i)=0$ for every $k$-algebra morphism $\varphi:A\to k$, so fix such a $\varphi$.
Now consider the map $\tilde{\varphi}:C\to k\otimes_k B\cong B$ induced by $(\varphi,\mathrm{id}_B)$. Then we have $\tilde{\varphi}(c)=\sum_{i\in[n]}\lambda_i\varphi(a_i)b_i$, and I claim that $\tilde{\varphi}(c)=0$. By hypothesis on $B$, it suffices to show that $\psi(\tilde{\varphi}(c))=0$ for every $k$-algebra morphism $B\to k$, so fix such a $\psi$. Then $\theta=\psi\circ\tilde{\varphi}$ gives a $k$-algebra morphism $C\to k$, and so by hypothesis on $c$ gives $\theta(c)=0$, as desired.
So $\tilde{\varphi}(c)=0$, which by linear independence of $\{b_1,\dots,b_n\}$ forces $\lambda_i\varphi(a_i)=0$ for all $i\in[n]$, and hence $\varphi(a_i)=0$ for all $i\in[n]$, as needed.

As a side remark, note that, if $k$ is algebraically closed and $A$ is finite type, then your condition is equivalent to reducedness.
