In these situations you need to you the distributivity laws, in your case you should proceed using
$$(a\wedge b)\vee c \leftrightarrow (a\vee c)\wedge (a\vee b), $$
where $$c:=(\neg R\wedge Q).$$
So the complete solution is as follows:
$$(P\ \land\ (\neg Q \lor\ R))\ \land\ ((\neg S\ \land\ \neg P)\ \lor\ (\neg R\ \land\ Q))$$
$$(P\ \land\ (\neg Q \lor\ R))\ \land\ ((\neg S\ \vee\ \neg R)\ \wedge\ (\neg S\ \vee\ Q)\ \wedge (\neg P\ \vee \neg R)\ \wedge (\neg P\vee Q))$$
To satisfy this formula it must be the case $v(P)=1$, then we must have $v(R)=0$ and $v(Q)=1$, therefore this valuation cannot satisfy the clause $\neg Q \vee R$.
Resolution: You can (using resolution) infer $\{Q,\neg R\}$ applying $P$ to $\neg P\vee Q$ and $\neg P\vee\neg R$. Finally $\{Q,\neg R\}$ and $\neg Q\vee R$ gives the set $\{\neg R,R\}$, which implies the formula cannnot be satisfiable.
Comment: If you want to show that a formula is tautology, I would not negate the original formula, just transform it into conjunctive normal form. The formula is then a tautology if and only if every every clause of the CNF contains both $A$ and $\neg A$ for some atom $A$.
Proof of the last claim: from right to left: consider any valuation $v$ then it obviously must satisfy every clause, because it contains some $A$ and $\neg A$. On the other hand if you have some clause where for every atom $A$ at least on of $A$ and $\neg A$ is not in the clause. It is enough to define $v(A)=0$ iff $A$ is in the clause. It easily follows that vv does not satisfy the clause and hence the whole formula. QED
It is true that, thanks to completeness (and soundness) theorem, proving Points 1 and 2 is equivalent to prove Points 3 and 4.
And your proof of Point 4 is correct.
But your proof of Point 3 is not correct, and actually Point 3 (and hence Point 1) does not hold!
Where is the error in your attempt to prove Point 3? From the fact that $\mathcal{A} \models \lnot \phi$ it does not follow that $\Sigma \models \lnot \phi$.
Indeed, $\Sigma \models \lnot \phi$ means that for every model $\mathcal{A}'$, if $\mathcal{A}' \models \Sigma$ then $\mathcal{A}' \models \lnot \phi$.
But in your attempt to prove Point 3, you have just shown that there exists a model $\mathcal{A}$ such that $\mathcal{A} \models \Sigma$ and $\mathcal{A} \models \lnot \phi$.
A priori, it is still possible that there is another model $\mathcal{B}\models \Sigma$ such that $\mathcal{B} \models \phi$, your argument does not exclude this possibility. And actually this is what actually happens!
Why Point 3 does not hold? Suppose that $\Sigma$ is the set of axioms defining a group (it can be easily expressed in first-order logic, see here), and that $\phi$ is the formula expressing that the group is abelian. Clearly, $\Sigma \not\models \phi$ because not all groups are abelian, but from that it does not follow that $\Sigma \models \lnot \phi$ because it is not true that all groups are non-abelian.
Best Answer
is a contradiction because it is of the form $\varphi \land \neg \varphi$
is a tautology, because it is of the form $\varphi \lor \neg \varphi$
Therefore, 2. is a tautology as well, since it is equivalent to 3
So all of that makes sense: If 1 is a contradiction, then 2, being the negation of 1, is a tautology.
Now, as far as formally demonstrating these things: in many systems pointing out that a statement is of the form $\varphi \land \neg \varphi$ is sufficient to prove that it is a contradiction. And in many systems, showing that a statement is of the form $\varphi \lor \neg \varphi$ is sufficient to show it is a tautology.
Some systems, however, have an explicit contradiction symbol, $\bot$, and so to show that a statement is a contradiction you either show that it is equivalent to $\bot$, or derive $\bot$ from it (any statement from which a contradiciton can be derived is a contradiction itself).
In some systems you can go straight to $\bot$ from $\varphi \land \neg \varphi$ as either a rule of equivalence or rule as inference, but in other systems you derive $\bot$ from a statement $\varphi$ and a second statement $\neg \varphi$
Likewise, some systems have an explicit symbol for the tautology: $\top$, and they will typically have an equivalence rule to go from $\varphi \lor \neg \varphi$ to $\top$. Unlike the $\bot$, though, formal systems typically do not have an inference rule to derive $\top$, because a tautology logically follows from any statement, and this the fact that a tautology can be inferred from some statement says nothing about that statement.
Instead, many systems will demonstrate a statement to be a tautology by demonstrating that its negation is a contradiction. This is the proof by contradiction proof technique of course.
Now, you actually do something very unusual: you negate statement 1, and show that the result is equivalent to a tautology. And yes, while that indeed show that the original statement is a contradiction, there would be few systems that would demonstrate statements to be a contradiction in this manner. As explained above, this would only work if you with equivalence rules only, and most such systems have a much more direct way to demonstrate a statement to be a contradiction.
Indeed, if you are talking about natural deduction, then you are probably talking about a system of inference, and remember that if you end up inferring a tautology from the negation of some statement, then that tells us nothing about the nature of that negation, and thus nothing about the original statement either.
So no, you typically do not demonstrate a statement to be a tautology using natural deduction by negating it and seeing what happens. Or, put differently, while a proof by contradiction demonstrates a statement to be a tautology by showing that the negation of that statement leads to a contradiction, in natural deduction there is no parallel proof technique that shows a statement to be a contradiction by showing that its negation leads to a tautology.