1 . This is Lagrange's theorem. If $G$ is the group $(\mathbb{Z}/13\mathbb{Z})^{\ast}$ (the group of units modulo $13$), then the order of an element $a$ (that is, the smallest number $t$ such that $a^t \equiv 1 \pmod{13}$) must divide the order of the group, which is $\varphi(13) = 12$. So we only check the divisors of $12$.
2 . Yes, that is a square mod $13$. To say that $a$ is a primitive root mod $13$ means that $a^{12} \equiv 1 \pmod{13}$, but all lower powers $a, a^2, ... , a^{11}$ are not congruent to $1$. Again use Lagrange's theorem: supposing $a^2$ were a primitive root, then $12$ would be the smallest power of $a^2$ such that $(a^2)^{12} \equiv 1$. But note that $b^{12} \equiv 1$ for ANY integer $b$ not divisible by $13$. So $(a^2)^{6} = a^{12} \equiv 1$, and $6 < 12$, contradiction.
3 . It's a general result about finite cyclic groups. A cyclic group of order $m$ is a group of the form $H = \{ 1, g, g^2, ... , g^{m-1}\}$. It is basically the same thing as the group $\mathbb{Z}/m\mathbb{Z}$ with respect to addition. In general, if $d \geq 1$, there exist elements in $H$ with order $d$ (that is, their $d$th power is $1$, all lower powers are not $1$) if and only if $d$ is a divisor of $m$, and there are exactly $\varphi(d)$ such elements.
In particular, if $p$ is an odd prime number, the result is that $(\mathbb{Z}/p\mathbb{Z})^{\ast}$ is a cyclic group of order $\varphi(p) = p-1$, and the number of primitive roots (that is, the number of elements with order $p-1$) is exactly $\varphi(p-1) = \varphi(\varphi(p))$.
4 . If you have found a primitive root modulo $p$ (where $p$ is an odd prime), then you can easily find the rest of them: if $a$ is a primitive root mod $p$, then the other primitive roots are $a^k$, where $k$ runs through those numbers which don't have any prime factors in common with $p-1$. It's a good exercise to prove this. So $2^9$ wouldn't work; $9$ has prime factors in common with $12$.
Best Answer
Let $g$ be a primitive root of $83$. Then all the primitive roots are $g^k$ where $k$ is relatively prime to $82$, so all $g^k$ with odd $k$ from $k=1$ to $k=81$, with the exception of $k=41$. These are all the quadratic non-residues of $83$ except $-1$.
Let $S$ be the sum of the primitive roots. Then the sum of the quadratic non-residues of $83$ is congruent to $S-1$ modulo $p$. This is $0$ modulo $83$, so $S\equiv 1\pmod{83}$.
The sum of the squares of the quadratic non-residues is congruent to $0$ modulo $83$. This is because $83$ is of the form $4k-1$, so $x$ is a QR if and only if $-x$ is an NR. So the sum $T$ of the squares of the primitive roots is congruent to $-1$.
Our sum of products is $\frac{1}{2}(S^2-T)$, which is congruent to $1$ modulo $p$.