As correctly said by Mauro curto in his comment, the missing step in your attempt of derivation is the use of the inference rule $\lor \mathbf{E}$ for eliminating the disjunction $p \lor q$.
The idea is that, because of the first premise $(p \lor q) \land (r \to \lnot p)$, the disjunction $p \lor q$ holds but it is unknown if $p$ holds or $q$ holds. In the first case, since $r \to \lnot p$, you can easily infer $\lnot r$ (via modus tollens). In the second case, $\lnot r $ immediately follows because of the second premise.
Therefore, a correct derivation in natural deduction of $\lnot r$ from the premises $(p \lor q) \land (r \to \lnot p)$ and $q \to \lnot r$ is the following:
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\oi#1{\qquad\mathbf{\lor I} \: #1 \\}
\def\oe#1{\qquad\mathbf{\lor E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\neg I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
$
\fitch{1. \, (p \lor q) \land (r \to \lnot p) \qquad \text{premise}
\\
2.\, q \to \lnot r \qquad \text{premise}
}
{ 3. \, p \lor q \ce{(1)}
\fitch{4.\, p \qquad \text{assumption}}
{ 5. \, r \to \lnot p \ce{(1)}
\fitch{6. \, r \qquad \text{assumption}}
{
7. \, \lnot p \ie{(6, 5)}
8. \, \bot \ne{(7, 4)}
}
\\
9. \, \lnot r \ni{(6{-}8)}
}\\
\fitch{10.\, q \qquad \text{assumption}}
{ 11. \, \lnot r \ie{(2, 10)}
}\\
12. \, \lnot r \oe{(3{-}11)}
}
$
Note that in your attempt of derivation, $p \lor q$ need not be assumed, because it follows from the first premise $(p \lor q) \land (r \to \lnot p)$ by means of the inference rule $\land \mathbf{E}$ for elimination of conjunction.
Best Answer
$$(¬p ∧ (p ∨ q)) → q \tag{given}$$
$$\equiv [\underbrace{(\lnot p \land p)}_{\bot} \lor (\lnot p \land q)] \to q\tag{distributive law}$$
$$\equiv \bot \lor (\lnot p \land q) \to q $$
$$\equiv (\lnot p \land q) \to q$$
$$ \equiv \lnot (\lnot p \land q) \lor q$$
$$\equiv (p \lor \lnot q) \lor q$$
$$\equiv p \lor (\lnot q \lor q)$$
$$p \lor \top$$
$$\top$$
Can you supply the reasoning here?