Proof of (P → Q) from ¬P?

The details of the proof will depend on the proof system you want to use.

• Clue 1: P → Q is classically equivalent to $ (\neg P \vee Q)$
• Clue 2: $ (\neg P \vee Q)$ is a dilution of $\neg P$

I’m going to answer on the basis of the derivational system suggested by the terminology that you’re using for your rules. What I’m doing may not be quite like what you’ve been taught, but I hope that it’s at least close.

You have $\neg P$ as hypothesis, and from that you want to derive $P\to Q$. To get $P\to Q$ by conditional introduction, you’ll need to start by taking $P$ as an assumption:

$$\begin{align*} &1.\quad \neg P\tag{Premise}\\ &2.\quad|\,P\tag{Assumption} \end{align*}$$

We want to end up concluding $Q$ within the scope of the assumption, since conditional introduction will then give us $P\to Q$. Introduce a second assumption layer:

$$\begin{align*} &1.\quad \neg P\tag{Premise}\\ &2.\quad|\,P\tag{Assumption}\\ &3.\quad||\,\neg Q\tag{Assumption}\\ &4.\quad||\,P\\ &5.\quad||\,\neg P\\ &6.\quad||\,\bot\\ &7.\quad|\,\neg\neg Q\\ \end{align*}$$

Here $\bot$ is a standard symbol for a contradiction, and deriving the contradiction from the assumption $\neg Q$ gives me $\neg\neg Q$ by negation introduction, thereby discharging the inner assumption; I’ll leave the other justifications so far to you.

Now you need some steps deriving $Q$ from $\neg\neg Q$; I’ll leave them to you. (You may have done or seen this derivation already.) Once those are in place, you can discharge the outer assumption by conditional introduction to get $P\to Q$.

The simplest way would be to use a truth table, which is just as rigorous if not as appealing. If you can't though then I'd consider the following.

$$\neg P \to Q \lor \neg P$$

and then using a simple variation of what is considered a definition by some texts $$Q \lor \neg P \iff \neg(\neg P)\to Q$$ implies what you want.