Given a finite dimensional selfinjective algebra A and an indecomposable nonprojective module N. Let $M:=A \oplus N$ and $B:=End_(M)$.
Does $B$ always has dominant dimension equal to the finitistic dimension?
The problem can be restated as follows: Let $d:=\inf \{ i \geq 1 | Ext^{i}(M,M) \neq 0 \} +1$ (this is the dominant dimension of B) and $M^{\perp d-2} := \{ X | Ext^{i}(M,X)=0 $ for $i=1,...,d-2 \}$. The question can be restated wheter every module in $M^{\perp d-2}$ but not in add(M) has infinite $add(M)$-resolution dimension.
It is true in the following cases:

1.$A$ local and $M$ with $Ext^{1}(M,M) \neq 0$, a proof can be found here http://mathoverflow.net/questions/257744/finite-addn-resolution/257839#257839 as an answer. (For example this included all local Hopf algebras)

2.$A$ serial and $N$ a simple module or a radical of a projective indecomposable module (I only have a highly complicated proof of that at the moment). It is also true for various other representation-finite symmetric algebras when choosing certain simple modules.

3.N with $\tau(N) \cong N$.

This would be highly surprising if always true, but the calculation of finitistic dimension is highly complicated and my computer can just test it for some small cases where B is representation finite. Had no luck with a counterexample yet.