(i) Prove that the last digit of $2^{2k}$ is either 4 or 6.
(ii) Prove that for every even perfect number the last digit is either a 6 or an 8.
Note the proposition is not true for $k=0$, so we proceed assuming $k>0$.
(i) The last digit of a number is congruent to that number modulo 10.
We use induction to prove the result. Let the statement $P(k)$ be the proposition
$$ P(k) := \quad 2^{2k} \equiv 4 \pmod {10} \quad \lor \quad 2^{2k} \equiv 6 \pmod {10} $$
We need to prove the base case $P(1)$, and the induction step $P(k) \implies P(k+1)$. Note the proposition is not true for $k=0$.
Base Case $P(1)$
Let's consider $k=1$
$$ 2^{2 \times 1} \equiv 2^2 \equiv 4 \pmod {10} $$
And so the base case is true.
Induction Step $P(k) \implies P(k+1)$
We assume the induction hypothesis $P(k)$.
Consider
$$ 2^{2(k+1)} \equiv 2^{2k} \times 2^2 \equiv 4N \pmod {10} $$
where $N = 2^{2k}$.
The induction hypothesis tells is that $N \equiv 4 \pmod {10}$ or $N \equiv 6 \pmod {10}$. Let's consider each case:
$$N \equiv 4 \pmod {10} \quad \implies \quad 4N \equiv 16 \equiv 6 \pmod {10}$$
$$N \equiv 6 \pmod {10} \quad \implies \quad 4N \equiv 24 \equiv 4 \pmod {10}$$
So in all cases, the induction step $P(k) \implies P(k+1)$ is true.
By induction we have shown that $2^{2k}$ is 4 or 6 modulo 10. That is, the last digit of $2^{2k}$ is either 4 or 6.
(ii) We recall that every even perfect number $N$ is of the form
$$N = 2^{p-1}(2^p-1)$$
where $(2^p-1)$ is prime, which also means $p$ is prime.
Case $p$ even
We first deal with the first even perfect number, $N=6$, which corresponds to $p=2$, the only even prime. In this case, the last digit is 6, and so the statement that the last digit is 6 or 8 is true.
Case $p$ odd
We now consider even perfect numbers corresponding to $p$ odd. In this case $p-1$ is even, and so $2^{p-1}$ can be written as $2^{2k}$ for some integer $k$. We've shown above that
$$ 2^{p-1} \equiv 2^{2k} \equiv 4 \pmod {10} \quad \lor \quad 2^{p-1} \equiv 2^{2k} \equiv 6 \pmod {10}$$
Let's consider each of these cases in turn.
For $2^{p-1} \equiv 6 \pmod {10}$
$$ \begin{align} 2^{p-1} & \equiv 4 \pmod{10} \\ \\ 2^p & \equiv 8 \pmod {10} \\ \\ 2^p - 1 & \equiv 7 \pmod {10} \\ \\ 2^{p-1}(2^p-1) & \equiv 4 \times 7 \pmod {10} \\ \\ & \equiv 8 \pmod{10} \end{align} $$
For $2^{p-1} \equiv 8 \pmod {10}$
$$ \begin{align} 2^{p-1} & \equiv 6 \pmod{10} \\ \\ 2^p & \equiv 12 \pmod {10} \\ \\ 2^p - 1 & \equiv 1 \pmod {10} \\ \\ 2^{p-1}(2^p-1) & \equiv 6 \times 1 \pmod {10} \\ \\ & \equiv 6 \pmod{10} \end{align} $$
We have shown that every even perfect number, which is of the form $2^{p-1}(2^p-1)$ where $(2^p-1)$ is prime and so $p$ is prime, has a last digit of 6 or 8.