1. really cant do

2. Hello,
For question (a): Each number n in the sum is lower than p. So, (n,p)=1.
Hence, we can apply Fermat's Little Theorem:
$n^{p-1}\equiv 1\text{ mod }p$
Let's multiply both sides by n.
$n^{p}\equiv n\text{ mod }p$
Thus
$\sum_{n=1}^{p-1} n^p\equiv \sum_{n=1}^{p-1} n\equiv \frac{(p-1)(p)}{2}\text{ mod }p$
Isn't it ?

3. For (a) we can strengthen the problem. $1^k + 2^k + ... + (p-1)^k \equiv 0 (\bmod p)$ if $(p-1)\not | k$ and $\equiv p$ if $(p-1)| k$.

For (b) note $1+...+p = p\cdot \frac{p-1}{2}$. Let $a=\frac{p-1}{2}$.
Let $r$ be remainder of $(p-1)!$ mod $pa$.
Thus, $(p-1)!\equiv r(\bmod pa)$ thus $(p-1)!\equiv r(\bmod p)$ and $(p-1)!\equiv r(\bmod a)$.

By Wilson's theorem it follows that $r\equiv -1(\bmod p)$.
Also $r\equiv 0(\bmod a)$ because $a$ divides $(p-1)!$.
If we let $r=p-1$ it satisfies both conditions.

4. For the second problem let $y$ be the order of $a$ mod $b$. Let $a^x\equiv 1(\bmod b)$. Write $x=qy+r$ where $0\leq r < y$. Then, $a^x \equiv 1 (\bmod b)\implies a^{qy+r}\equiv 1(\bmod b)\implies a^r\equiv 1(\bmod b)$. Since $y$ is the order of $a$ it means $a^r\not \equiv 1$ if $0. But since $0\leq r < y$ and $a^r\equiv 1(\bmod b)$ it follows that $r=0$. And thus $x=qy+r = qy\implies y|x$.