Proving the remainder is $1$ if the square of a prime is divided by $12$

$ p^2 -1 =(p-1)(p+1) $. Since $p$ is odd, these are two even factors, so the product is divisible by $4$. Also, one of $p-1$, $p$, $p+1$ must be divisible by $3$, but since $p$ is prime, it is not divisible by $3$. Thus one of $p-1$, or $p+1$ must be divisible by three.

Hint: consider $0,1,2,\ldots, 11$, which are all the possible remainders modulo 12. Take out the ones that couldn't be congruent to a prime bigger than 3. Square all the ones that are left modulo 12, and see what you get.

Hint: $\,3 < p\,$ prime $\,\Rightarrow\, p = 6n\pm 1\,\Rightarrow\, p^2 = 36n^2\pm12n+1$