Solution:
The answers are $3$, $8$, and $0$.

It is clear that $p=2$ gives $3$, $p=3$ gives $8$, and $p=5$ gives $0$. We claim that all primes $p \geq 5$ give the remainder $0$ as well, i.e. that $p^{2}-1$ is divisible by $12$ for these $p$.

We factor:
$$p^{2}-1=(p+1)(p-1)$$
Since $p \neq 2$, $p$ is odd and so $p+1$ and $p-1$ are both even. This gives two factors of $2$ in $p^{2}-1$. Moreover, one of the three consecutive integers $p-1$, $p$, $p+1$ is divisible by $3$, and since $p \neq 3$, it is not $p$. So either $p-1$ or $p+1$ has a factor of $3$, and so does $p^{2}-1$. Thus $p^{2}-1$ is divisible by $2 \cdot 2 \cdot 3=12$.