(a) Show that if $a$ is an even number and $n$ is a natural number then $a^n$ is also even.
(b) Show that if $a$ is an odd number and $n$ is a natural number then $a^n$ is also odd.
We'll be using Proposition (3.8) that
$$a \equiv b \pmod m \implies a^k \equiv b^k \pmod m$$
(a) If a number is congruent to 0 modulo 2 then it is even.
Using Proposition (3.8),
$$ a \equiv 0 \pmod{2} \implies a^n \equiv 0^n \equiv 0 \pmod{n}$$
That is, $a^n$ is even.
Note that $n$ must be greater than 0.
(b) If a number is congruent to 1 modulo 2 then it is odd.
Using Proposition (3.8),
$$ a \equiv 1 \pmod{2} \implies a^n \equiv 1^n \equiv 1 \pmod{n}$$
That is, $a^n$ is odd.