, 2022

Solution:

Firstly, $f(x)$ and $g(x)$ must intersect - otherwise, $f(x) = g(x) = 1$, which can't be true.

Secondly, suppose they intersect at $a$, so that $f(a) = g(a) = c$. Then $f(c) = g(c) \Longrightarrow f(c) = g(c) = 1$. But then, $a = c$, and $c = 1$. So $f(1) = 1$, and we're done.

Solution:

We will solve the problem manually, setting $f(x) = a x + 2022$ and $g(x) = -a x - 2020$. Then $f(f(x)) = a^{2} x + 2022 a + 2022$, while $g(g(x)) = a^{2} x + 2020 a - 2020$.

Then, $a = -2021$, so then $f(1) = 1$.