2. Note that $\displaystyle 1^2\equiv (p-1)^2(\bmod.p)$, $\displaystyle 2^2\equiv (p-2)^2(\bmod.p)$, ...
So that the only distinct quadratic residues are: $\displaystyle 1^2(\bmod.p)...\left(\tfrac{p-1}{2}\right)^2(\bmod.p)$, but we have $\displaystyle p-1=|\mathbb{Z}_p^{\times}|$ hence the rest of them must be non-quadratic residues, that is we have $\displaystyle \tfrac{p-1}{2}$ quadratic residues and $\displaystyle \tfrac{p-1}{2}$ non-quadratic residues.
EDIT: Thought I'd clarify a bit: $\displaystyle x^2\equiv{y^2}(\bmod.p)$ if and only if $\displaystyle (x-y)\cdot (x+y)\equiv{0}(\bmod.p)$ since p is prime either $\displaystyle x \equiv{y}(\bmod.p)$ or $\displaystyle x \equiv{p-y}(\bmod.p)$