Q:

Solve for $f(x)$ and $g(x)$

we have to find all the solutions of $f(x)$ and $g(x)$, given that,
$f(x)\cdot g(x) = f(x) + g(x)$
$(f(x) - 1 )\cdot( g(x) - 1 ) = 1$
I have found out 2 solutions so far, $f(x)= g(x) = 2$, and $f(x) = \sec^2(x)$ , $g(x) = \csc^2(x)$, (and vice versa)

A:

The two equations you gave are actually just forms of the same information, since $1= ( f-1)(g-1) = fg - f -g + 1 .$ These are infinitely many solutions  to this for general functions $ f: \mathbb{R} \to \mathbb{R} $, or even if we assume stronger conditions of the functions like continuity, differentiability etc.  
For any function $ f: \mathbb{R} \to \mathbb{R} $ where $f(x)\neq 1$ for all $x\in \mathbb{R}$ , we can pair it with $\displaystyle g(x) = \frac{1}{ f(x) -1} +1 $ to form a solution. 