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.
We know that $a^{\phi(m)}\equiv1$ mod $m$ if $(a,m)=1$ and $a^{\phi(n)}\equiv1$ mod $n$ if $(a,n)=1$. Now let $L=\text{lcm}(\phi(m),\phi(n))$. Then $a^L\equiv1$ mod both $m$ and $n$ if $(a,mn)=1$. So if $(m,n)=1$, then $a^L\equiv1$ mod $mn$ for all $(a,mn)=1$. Finally, if $m,n\gt2$, then $2$ divides both $\phi(m)$ and $\phi(n)$, hence $L\le\phi(m)\phi(n)/2=\phi(mn)/2$ (the final equality because $m$ and $n$ are relatively prime). So there is no element of order $\phi(mn)$, which is to say there is no primitive root.
Remark: This is essentially the same proof as in lhf's answer (which I didn't see until I finished posting).
Best Answer
The have to be relatively prime to $n$ because if $x$ is a primitive root, then $x^n \equiv 1$ for some $n$, and therefore $x^{n-1}$ must be the multiplicative inverse of $x$. But only numbers relatively prime to $n$ have a multiplicative inverse.
They don't strictly speaking have to be smaller than $n$, but any number larger than $n$ is equivalent modulo $n$ to some number smaller than $n$. Remember that $$ a \equiv b \mod n \quad\Leftrightarrow\quad \exists k \in \mathbb{Z} \,:\, a - b = kn \text{.} $$ So if $a \geq n$, we can substract $n$ until we reach an integer $b$ within $[0,n-1]$. If we had to subtract $k$ times, we have $b = a - kn$, i.e. $a - b = kn$, so $a \equiv b \mod n$.