1) Let $p=4k+1$. Since $r$ is a primitive root of $p$, it follows that $r$ is a quadratic non-residue of $p$. Thus by Euler's Criterion, we have $r^{2k}\equiv -1\pmod{p}$.
It follows that $(-r)^{2k}\equiv -1\pmod{p}$. Multiply by $-r$. We get that $(-r)^{2k+1}\equiv r\pmod{p}$.
Since $r$ is a power 0f $-r$, and every $a$ relatively prime to $p$ is congruent to a power of $r$, it follows that every $a$ relatively prime to $p$ is congruent to a power of $-r$. This implies that $-r$ is a primitive root of $p$.
2) The result is easy to prove for $n=1$ and $n=2$, so we may assume that $n\ge 3$. Let $g$ be a primitive root of $n$. Then the $\varphi(n)$ numbers in the interval from $1$ to $n$ that are relatively prime to $n$ are congruent (modulo $n$) in some order to $g^1,g^2,g^3,\dots, g^{\varphi(n)}$.
Thus their product is congruent to $g^N$, where by the usual formula for the sum of the first $\varphi(n)$ positive integers we have $2N=\varphi(n)(\varphi(n)+1)$. Since $n\ge 3$, the number $\varphi(n)$ is even, so $\varphi(n)+1$ is odd.
We have $g^N=(g^{\varphi(n)/2})^{\varphi(n)+1}$. Since $g$ is a primitive root of $n$, we have $g^{\varphi(n)/2}\equiv -1\pmod{n}$. Raising this to the odd power $\varphi(n)+1$, we see that $g^N\equiv -1\pmod{N}$, which is what needed to be shown.
Hint: if $(n, p-1)=1$ then you can find $a, b$ such that $an+(p-1)b=1$. Then for any $x$ invertible mod $p$,
$$(x^n)^a = x^{an} = x^{an+(p-1)b} = x.$$
(I used that $x^{p-1}=1$.)
Therefore the map $x \mapsto x^n$ has a double sided inverse given by $y \mapsto y^a$...
Best Answer
For $(a),$
If $g$ is a primitive root, all positive integers not exceeding $m$ and relatively prime to $m$ will be $g^r, 1\le r\le \phi(m)$
So, the required product $$\equiv \prod_{1\le r\le \phi(m)}g^r=g^{\sum_{1\le r\le \phi(m)}r}=g^{\frac{\phi(m)(\phi(m)+1)}2}=\left(g^{\frac{\phi(m)}2}\right)^{\phi(m)+1}$$
Using this, $g^{\frac{\phi(m)}2}\equiv-1\pmod m$ as $g^{\frac{\phi(m)}2}\not\equiv1\pmod m$ as $g$ is a primitive root
and we know $\phi(m)$ is even for $m\ge3\implies\phi(m)+1$ is odd