Question. 7 show P, ¬Q: ¬ (P →Q) (should be doable in 6 steps)

You tried

{P}(1) P Premise

{¬P}(2)¬P Premise

{¬Q→P ?}(3)¬Q→P ? 1

{¬¬Q?}(4)¬¬Q by MT 2, 3

{Q}(5)Q 5. DNE ?

⋮ ?

Not horrible, but it could use some improvement.

All right, first thing some people do with any formal proofs, is list out your assumptions and desired conclusion. It lets you know what you have to work with, and where you want to go. In this case we're lucky and have an idea how long it should be too so we can put the last line number in too, ordinarily you don't know it until the proof is complete.

1(1)P 2(2)¬Q ⋮ 1,2(6)¬(P→Q)

Now here is where you have to step back and think about what it is you want to do, and how you might be able to do it. Recall the Tomassis advice

Each of these proofs requires that you first derive the relevant conditional and then apply MT.

So it's more than a good bet MT will feature prominently.

In each case, you will have to derive the conditional by augmenting the premises.

Good to keep in mind, and

In these cases it is crucial to identify clearly the formula you ultimately want to derive, i.e. the conclusion of the particular sequent you are trying to prove.

Ok. So we have a pretty good idea we want to use MT, and might have to assume something to augment the premises. So how about seeing what you'd need if you have

1(1)P 2(2)¬Q ⋮ (5) ?? → ???
1,2(6)¬(P→Q) MT 5,?

You should be able to fill in at least half the conditional in (5) pretty easily. Just consider what (5) needs to look like to conclude (6). That's half. The other half is almost as straightforward. I can give you a hint if you get really stuck and frustrated, but see what you can do first.