It never hurts to gather some numerical data. For your first problem:
$$\begin{array}{cc|cc|cc} n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n&n&\left(\sum_{k=1}^{n-1}k^2\right)\bmod n\\ \hline 2&1&8&4&14&7\\ 3&2&9&6&15&10\\ 4&2&10&5&16&8\\ 5&0&11&0&17&0\\ 6&1&12&2&18&3\\ 7&0&13&0&19&0 \end{array}$$
Examination of that table should suggest that
$$\left(\sum_{k=1}^{n-1}k^2\right)\bmod n=\begin{cases} \frac{n}2,&\text{if }n\equiv 2\!\!\!\pmod 6\\\\ \frac{2n}3,&\text{if }n\equiv 3\!\!\!\pmod 6\\\\ \frac{n}2,&\text{if }n\equiv 4\!\!\!\pmod 6\\\\ 0,&\text{if }n\equiv 5\!\!\!\pmod 6\\\\ \frac{n}6,&\text{if }n\equiv 0\!\!\!\pmod 6\\\\ 0,&\text{if }n\equiv 1\!\!\!\pmod 6\;; \end{cases}\tag{1}$$
since you know that $$\sum_{k=1}^{n-1}k^2=n\left(\frac{n^2}3-\frac{n}2+\frac16\right)=\frac16n(n-1)(2n-1)\;,$$ it shouldn’t be too hard to prove $(1)$.
Added: For the second problem, did you try calculating
$$\left(\sum_{k=1}^{n-1}k^3\right)\bmod n=\frac{n^2(n-1)^2}4\bmod n$$
for $n=2,3,4,\dots$ as I did for the first problem? Doing so should let you guess right away that the value is $0$ for odd $n$, and discovering the mathematical reason for this isn’t hard. If $n$ is odd, then $\frac{(n-1)^2}4$ is an integer, and clearly $n^2\frac{(n-1)^2}4\bmod n=0$.
If $n$ is even, then
$$\frac{n^2(n-1)^2}4\equiv\left(\frac{n}2\right)^2(n-1)^2\equiv\left(\frac{n}2\right)(-1)^2\equiv\left(\frac{n}2\right)^2\pmod n\;.$$
Now make a table for $n=2,4,6,\dots$:
$$\begin{array}{rccc} n:&2&4&6&8&10&12&14&16\\ \hline \left(\frac{n}2\right)^2\bmod n:&1&0&3&0&5&0&7&0 \end{array}$$
The pattern is pretty obvious: it definitely appears that
$$\left(\frac{n}2\right)^2\bmod n=\begin{cases} 0,&\text{if }n\equiv 0\!\!\!\pmod 4\\\\ n/2,&\text{if }n\equiv 2\!\!\!\pmod 4\;, \end{cases}$$
and you shouldn’t have too much trouble proving that this is actually the case.