I believe 2 is equivalent to 3. First, I'll give a proof for fields (added then I will apply the crux of the argument to the general case). Suppose that $K$ is an infinite extension of $k$. Then $K\otimes_k K$ is not Noetherian and hence not Artinian semisimple (by Hopkins-Levitzki). This follows from Theorem 11 of P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35.
Update: Here is a proof that 2 is equivalent to 3 that reduces to the argument above for a field. Let $A$ be an infinite dimensional $k$-algebra which satisfies 2. We get a contradiction. Note $A$ must be semisimple so it is a finite direct product of matrix algebras over division algebras and one of those division algebras $D$ must be infinite dimensional over $k$. Since tensor product commutes with direct product and plays nicely with matrices, it suffices to show that $D\otimes_k K$ is not separable for some field $K$. Note that $Z(D)$ is a field. Suppose first that $D$ is finite dimensional over $Z(D)$. Then $Z(D)/K$ is an infinite extension and so by the field case $Z(D)\otimes_K Z(D)$ is not Noetherian (see above). But $Z(D\otimes_k K) \cong Z(D)\otimes_K K$ by standard stuff about centers of tensor products of algebras over a field. Therefore, $D\otimes_k K$ is not semisimple because the center of a semisimple ring is a finite direct product of fields, hence Noetherian. So we may assume that $D/Z(D)$ is infinite.
I claim that $D$ contains subfields $F$ containing $Z(D)$ of arbitrary high dimension. Indeed, suppose there is $d$ so that $[F:Z(D)]\leq d$ for all $Z(D)\leq F\leq D$ with $F$ a subfield. If $\alpha\in D$, then $Z(D)(\alpha)$ is a subfield and so by assumption has degree at most $d$. But then $\alpha$ is the root of a polynomial over $Z(D)$ of degree at most $d$. It follows by a theorem of Jacobson-Kaplansky that $D$ is finite dimensional over $Z(D)$ (see Theorem 4 of I. Kaplansky, Rings with a polynomial identity. Bull. Amer. Math. Soc. 54 (1948), 575–580). Thus there are subfields $F$ of $D$ containing $Z(D)$ of arbitrary dimension. Then $Z(D\otimes_k F)=Z(D)\otimes_k F$ is not noetherian and hence not a direct product of fields. Thus $D\otimes_k F$ is not artinian.