Show that (p ∧ q) → (p ∨ q) is a tautology?

It is because of the following equivalence law, which you can prove from a truth table: $$r\rightarrow s\equiv \lnot r\lor s.$$ If you let $r = p\land q$ and $s = p\lor q$, you get what you are looking for, namely that $$(p\land q)\rightarrow (p\lor q)\equiv \lnot(p\land q)\lor(p\lor q).$$

This is a classic equivalence $$p\to q \equiv \neg p\lor q.\tag 1$$

We can examine equivalence using a truth table \begin{array}{c|c|c|c|c} p&q&\neg p&p\to q&\neg p\lor q\\\hline T&T&F&T&T\\\hline T&F&F&F&F\\\hline F&T&T&T&T\\\hline F&F&T&T&T \end{array}

Hence, $p\to q\equiv \neg p\lor q$.

Further, if $a \equiv p\land q$ and $b\equiv p\lor q$, and so
$$a\to b\equiv \neg a\lor b\equiv \neg(p\land q)\lor(p\lor q)\equiv (\neg p\lor \neg q)\lor (p\lor q),$$ by using $(1)$.

To show (p ∧ q) → (p ∨ q).

If (p ∧ q) is true, then both p and q are true, so (p ∨ q) is true, and $T \to T$ is true.

If (p ∧ q) is false, then (p ∧ q) → (p ∨ q) is true, because false implies anything.

Q.E.D.