In natural deduction proofs, it is good to "think ahead", and at any stage of a proof look at what you are trying to derive at that stage, and ask yourself "How can I prove that?"
So, in this case, you start with the premiss
$(P \to (Q \to R))$
and your target is to derive
$((P \land Q) \to R)$.
OK: ask what kind of proposition is that target? what is its main connective?? It's a conditional (with antecedent $(P \land Q)$ and consequent $R$). And how can you derive a conditional? Typically by using ImpIntro, yes? So the obvious thing to try is to assume the antecedent of the target and try to get to the consequent. So the proof is going to look like this ...
$(P \to (Q \to R))$
$\quad|\quad(P \land Q)$
$\quad|\quad\ldots$
$\quad|\quad R$
$((P \land Q) \to R)$.
Or at least, that's how it will look if we follow best practice and indent a proof every time we make a new assumption and finish the indented part when the assumption is "discharged" by the rule ImpIntro (i.e. the assumption is no longer in play).
So now our target is to get to $R$, and in this case we are allowed to use every above. It should be quite obvious how to do this.
But in case not, let's plod through, thinking strategically again. Where is $R$ to be found in the premisses? As the consequent of a conditional inside the consequent of a bigger conditional. How can we dig stuff out of a conditional? By modus ponens of course (what else)? How can we get to use modus ponens on a conditional? By having the antecedent to appeal to. So we need the antecedents $P$ and then $Q$ to get us to $R$. How can we get them? ... OK now you can see what to do to complete the proof for yourself!
Practice on some natural deduction examples and this kind of strategic thinking will become second nature!
For another example, let's go the other way about and try to get from the premiss
$((P \land Q) \to R)$
to the conclusion
$(P \to (Q \to R))$.
Again the conclusion is a conditional, so what do you do? You are going to use ImpIntro, so assume the antecedent and aim for the consequent:
$((P \land Q) \to R)$
$\quad|\quad P$
$\quad|\quad\ldots$
$\quad|\quad (Q \to R)$
$(P \to (Q \to R))$.
Now your aiming for the conditional $(Q \to R)$. How do you prove that? Assume the antecedent and derive the consequent of course. So let's do that ....
$((P \land Q) \to R)$
$\quad|\quad P$
$\quad|\quad|\quad Q$
$\quad|\quad|\quad\ldots$
$\quad|\quad|\quad R$
$\quad|\quad (Q \to R)$
$(P \to (Q \to R))$.
Note we have indented again, because we have a new temporary assumption in play, made while the previous one is still active. So now we need to derive $R$. How are we going to do that? Obviously we need to use the initial conditional premiss plus modus ponens. So we are going to need to derive the antecedent of that initial conditional, i,e, $(P \land Q)$. So .... we need
$((P \land Q) \to R)$
$\quad|\quad P$
$\quad|\quad|\quad Q$
$\quad|\quad|\quad\ldots$
$\quad|\quad|\quad(P \land Q)$
$\quad|\quad|\quad R$
$\quad|\quad (Q \to R)$
$(P \to (Q \to R))$.
But we are done! There is nothing more to be filled in (why?). Easy eh? The proof writes itself, if you think strategically.
Usually natural deduction proofs are easiest to construct from the bottom up. Whenever you need to prove something of the form $\varphi\to\psi$, your options are either to produce it using ${\to}E$ on $\sigma\to\varphi\to\psi$ and $\sigma$, or to produce it using ${\to}I$ on $\psi$. The first of these options requires some cleverness in choosing a formula $\sigma$; the second is more or less mechanical.
Whenever you use ${\to}I$ to conclude $\varphi\to\psi$, you get to state $\varphi$ as an assumption in the proof of $\psi$. And these are the only assumtions you ever get to make, at least in most orthodox formulations of Natural Deduction.
So if you're writing a proof from the top down, you have better not make any assumptions that you don't plan to "discharge" using ${\to}I$ further down in the tree.
In your second example is looks like you have the ${\lor}I$ rule wrong. It ought to be just
$$ \frac{P}{P\lor R} {\lor}I $$
with no $R$ premise. And that's good because you cannot assume $R$ in this place -- there is no instance of ${\to}I$ in the tree that concludes something of the form $R\to\cdots$.
In the last example, you;re trying to conclude $((P\to Q)\land (R\to Q)) \to (P \land R) \to Q$. Here the natural approach will be to end the proof with two stacked ${\to}I$s to produce the two arrows. Then your task is to conclude $Q$, while being allowed to assume $(P\to Q)\land (R\to Q)$ and $P\land R$. That's easy enough; would have been slightly more interesting of one of the $\land$s (but not both) had been $\lor$.
Best Answer
It's easier to show $\wedge$ distributes over $\vee$ than vice-versa. And the latter follows straightforwardly from the former. So you might first try showing $(Q \vee R) \wedge P \vdash (Q \wedge P) \vee (R \wedge P)$, and use those arguments or the fact itself in the needed proof. To show $\wedge$ distributes over $\vee$, make use of $P \wedge Q \vdash R$ if and only if $P \vdash Q \rightarrow R$. (This shows that $ \_ \wedge Q$ is a left adjoint of $Q \rightarrow \_$, and left adjoints preserve joins. For "and" to distribute over "or" means the same thing as the "and" operator to preserve binary disjunction, a join.) In formal deduction of $(Q \vee R) \wedge P \vdash (Q \wedge P) \vee (R \wedge P)$, you can use the inference law I mentioned to isolate $Q \vee R$ to the left of $\vdash$ and deal with $Q$ and $R$ separately. Even if you want to show the needed result directly, the idea is that whenever you need to show an inference with a hypothesis that is a conjunction where one conjunct is a disjunction, use the inference law I mentioned to create an equivalent inference with hypothesis the disjunction, which inference can be broken up into simpler inferences each involving as hypothesis a disjunct, and each of these inferences can be dealt with by using the reverse direction of the inference law to remove the implication that was created.