Suppose $A$ is commutative local (necessarily artinian) with the only simple $k\neq A$ then $\psi_k=1$, so your statement 1 will say that $Ext^1_A(M,M)=0$ implies $Ext^i_A(M,M)=0$ for all $i>0$. I stated it as a conjecture for complete intersections here (conjecture 9.1.3).

Note that if you combine it with a special case of the Auslander-Reiten conjecture, then we will have a stronger statement that $Ext^1_A(M,M)=0$ implies $M$ is free (still assuming that $A$ is Gorenstein). It was stated as a question (9.1.4) in the same survey.

As far as I know, both are open even for complete intersections unless $A$ is a hypersurface (which will be representation finite anyway).

Suppose $A$ is commutative local (necessarily artinian) with the only simple $k\neq A$ then $\psi_k=1$, so your statement 1 will say that $Ext^1_A(M,M)=0$ implies $M$ is free. I stated it as a conjecture for complete intersections here (conjecture 9.1.3). Technically, it was stated as $Ext^1_A(M,N)=0$ implies $Ext^i_A(M,N)=0$ for all $i>0$, but when $M=N$ the latter condition is equivalent to $M$ being free.

One could also ask if $Ext^1_A(M,M)=0$ implies $M$ is free, still assuming that $A$ is Gorenstein. It was stated as a question (9.1.4) in the same survey.

As far as I know, both are open even for complete intersections unless $A$ is a hypersurface (which will be representation finite anyway).