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.