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)