I believe 2 is equivalent to 3. I have a proof for fields but am still working on the details for 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.

More generally, if $A$ is commutative and infinite dimensional over $k$, then it doesn't satisfy 2 because if $A$ is semisimple (which it must be to satisfy 2), then it is a direct product of fields and one of those must be infinite dimensional over $k$.  Now use that the tensor product commutes with direct product and the field case.  

The general case reduces to a division algebra $D$ with $D$ infinite dimensional over $k$ satisfying $2$.  If $Z(D)/k$ is infinite, then $D\otimes_k Z(D)$ has center $Z(D)\otimes_k Z(D)$ which is not Noetherian (and hence not artinian) by the argument above and so $D\otimes_k Z(D)$ is not semisimple since the center of a semisimple ring is semisimple.  I don't yet know how to handle the case $Z(D)/k$ is finite.  Note that we may assume that $k=Z(D)$ since if $D$ satisfies 2 over $k$ it satisfies it over $Z(D)$ because $D\otimes_k K\to D\otimes_{Z(D)} K$ is surjective for any field extension $K$ of $Z(D)$ and semisimple algebras are closed under quotients.  So if you can prove 2 for D/Z(D) implies that $[D:Z(D)]<\infty$, then we finish (since $Z(D)/k$ is finite).