For how many primes $p$ is $p^2 + 2$ is also prime? For how many primes $p$ is  $p^2 + 2$ also prime?

I'm not sure but I think you can consider every prime number as $(3k+1)$ or $(3k+2)$


We will not consider $p=2$ because $2^2+2=6$ which is obviouly not prime.

Most definitely $p\neq n^2 + 1$ for all odd $p$ because if we assume otherwise, we want $$(n^2+1)^2+2=n^4+2n^2+3$$ to be prime. Therefore $3\nmid n$. (The only prime divisible by $3$ is $3$ itself, so $n=0$ is a solution, but note that $0^2+1=1$ which is neither prime nor composite.) Every prime is of the form $3k+1$ or $3k+2$, so if $p=n^2+1$ then $p\neq 3k+1$ and thus $p= 3k+2$. $$\begin{align}\therefore n^2+1&=3k+2 \\ \Leftrightarrow 3k&=n^2-1 \\ &=(n+1)(n-1).\end{align}$$ Since $$(n+1)-(n-1)=2<3$$ then $3\mid n+1$ or $3\mid n-1$ but not both. $$\therefore n = 3l+1\text{ or } 3l-1$$ which when we substitute, we get that $3\mid p^2+2$ for all odd $p = n^2+1$.

It is also provable by induction that $3\mid n^4 + 2n^2$ for all integers $n$. The same result is thus implied.
