Your point 5 is correct: given that $P \land Q$ eventually leads to a contradiction, you can conclude that $P \land Q$ is not true.
However, that does not invalidate any of the inferences you made from the assumption that $P \land Q$. That is, from the assumption that $P \land Q$, you can still infer $P$ as well as $Q$, and given those, you can respectively infer $R$ as well as $\neg R$ is true. And hence, you can infer a contradiction. All those inferences are still correct. And hence you can still say that "If $P \land Q$ is true, then $P$, $Q$, $R$, $\neg R$, and $\bot$ (logic symbol for contradiction) are all true as well". But, of course, since a contradiction can never be true, it follows that $P \land Q$ cannot be true either.
OK, so far so good. You also said that you were trying to prove $(P \land Q) \to R$. And, once again, you assumed $P \land Q$, inferred $P$, and from $P$ you were able to infer $R$. At that point, you can indeed infer $(P \land Q) \to R$. Once again, you have basically shown that "If $P \land Q$ is true, then $R$ is true as well", and by conditional proof we can wrap that up with $(P \land Q) \to R$.
Now (and I think this is your real/main question): does the fact that you can also infer $Q$, and from that $\neg R$, and thus a contradiction, take anything away from that? No. It is still true that "If $P \land Q$ is true, then $R$ is true as well". So, when you do a conditional proof, there is really no need to see if your assumption leads to a contradiction.
In fact, if some assumption does lead to a contradiction, then anything follows from your assumption, since anything follows from a contradiction. Hence the conditional that has the assumption as its antecedent ('if' part) will definitely be the case, no matter what you put down for the consequent ('then' part)
As you have proved, $\{\phi,\psi\}\vdash\psi\land\phi$ (since $\{\phi,\psi\}=\{\psi,\phi\}$ and according to point 3 of your exercise). And according to point 4 this means that $\{\phi\land\psi\}\vdash\psi\land\phi$. Using the deduction theorem we gain $\vdash(\phi\land\psi)\rightarrow(\psi\land\phi).$ Similarly, $\vdash(\psi\land\phi)\rightarrow(\phi\land\psi)$.
Now it's quite simple to conclude $\vdash(\phi\land\psi)\leftrightarrow(\psi\land\phi).$
Best Answer
No, because the sufficient property could be false: if $n=p=\top,s=\bot$, then $p⇒ n$ and $s⇒p$ are both true, but $n\wedge s=\bot$ while $p$ remains true, so the equivalence certainly cannot hold.
One could see that on an intuitive level by noting that since $s\implies p\implies n$, the statement $s\wedge n$ only depends on the statement of the stronger of both, namely the sufficient condition $s$. But that would reduce the right side to $s\Leftrightarrow p$, which is absurd in general.