For your second question, solving $39x \bmod 680 = 1\tag{1}$
is equivalent to solving the following congruence equation, $\bmod(680)$:
$39x \equiv 1 \pmod{680}.\tag{2}$
There is more than one solution: there are infinitely many solutions for $x$. Every integer $x$ which satisfies the following equation is a solution: $39x = 680k + 1$
Experiment with particular values for $k$ and see what values of $x$ you arrive at. Then try to define the set of all solutions.
ADDED: Solving $(2)$ gives us
$x \equiv 279 \pmod{680}.\tag{3}$
Then assuming you are looking for all integer solutions for $x$ we have, as solutions, all $x$ satisfying
$x = 680k + 279\quad k\in \mathbb{Z}.\tag{4}$
Note that when $k=0$, $x = 279$, which is the least positive solution solving your equation. So the set of all integer solutions satisfying $(1)$is given by $\{x\mid x =279 \pm 680k, k\in \mathbb{Z}\}.$
Please, in the future, if you have more than one sufficiently unrelated questions, post them separately.