Derived equivalence between two exotic algebras

Let $$A$$ and $$B$$ be two connected finite dimensional quiver algebras having the same underlying quiver.

Question 1:

In case $$A$$ and $$B$$ have exactly one indecomposable projective non-injective $$A$$-module and both have finite global dimension. Can they be derived equivalent in case they are non-isomorphic?

Probably the answer is yes and I just miss a good example.

Question 2:

In case $$A$$ and $$B$$ are Nakayama algebras that have exactly one indecomposable projective non-injective $$A$$-module and both have finite global dimension. Can they be derived equivalent in case they are non-isomorphic?

I think the answer to question 2 is no (it is no for algebras with $$n \leq 11$$ simple modules) but I could not find a good reason. There are exactly $$n$$ such Nakayama algebras with $$n$$ simple modules but their Kupisch series is surprisingly complicated to describe.

Here an example: The Nakayama algebras with Kupisch series [ 2, 2, 2, 2, 2, 3 ] and [ 3, 3, 3, 3, 4, 4 ] have both finite global dimension and the same Coxeter polynomial. But they are not derived equivalent since their fourth Hochschild cohomology is different.

Question 3:

Assume $$A$$ is an algebra having exactly one indecomposable projective non-injective $$A$$-module and dominant dimension at least one. In case $$A$$ has finite finitistic dimension $$d$$, the only tilting $$A$$-modules are $$eA \oplus \Omega^{-r}(A)$$ for $$r=0,1,…,d$$ when $$eA$$ is the minimal faithful projective-injective $$A$$-module. Can we even classify all tilting complexes of $$A$$ when it has finite finitistic dimension ?

(probably no, but maybe this is possible in some special cases)