We make extensive use of the identity $(a \to b) \equiv (\lnot a \lor b)$, and leave you to fill in the reasons for some of the intermediate steps in (2).
(1) $\quad p \to (q \wedge r) \equiv \lnot p \lor (q \land r) \equiv (\lnot p \lor q) \land (\lnot p \lor r) \equiv (p \to q) \wedge (p \to r)$.
(2) $\quad(p \to q) \land (p \lor q) \equiv q\quad?$ $(p \to q) \land (p \lor q) \equiv (\lnot p \lor q) \land (p \lor q)\tag{1}$ $\equiv \;[(\lnot p \lor q) \land p] \lor [(\lnot p \lor q) \land q]\tag{2}$ $\equiv \;[(\lnot p \land p) \lor (q \land p)] \lor [(\lnot p \land q) \lor (q \land q)]\tag{3}$ $\equiv \;F \lor (p \land q) \lor [(\lnot p \land q) \lor (q \land q)]\tag{4}$ $\equiv \;(p\land q) \lor [(\lnot p \land q) \lor (q \land q)]\tag{5}$ $\equiv \;(p\land q) \lor (\lnot p \land q) \lor q\tag{6}$ $\equiv \;[(p \lor \lnot p) \land q] \lor q\tag{7}$ $\equiv \;(T \land q) \lor q\tag{8}$ $\equiv \;q\lor q\tag{9}$ $\equiv \;q\tag{10}$