 Solution

Each nonzero quadratic residue is attained twice as $x^2$ for $1\le x\le p-1$. Therefore twice the sum of those residues is ${\sum }_{x=1}^{p-1}{x}^2$, which is zero modulo $p$ by (ii), since $p-1>2$. As two is invertible modulo $p$, the sum of the residues is zero. The sum of all nonzero residues is $p(p-1)/2\equiv 0$, so subtracting gives the same result for the nonresidues. Hence $\boxed{\text{both indicated sums are }0(\mathrm{mod}p)}$.