Solutions

a) Suppose there exist functions $f$ and $g$ satisfying $f(x)+g(y)=xy$ for all $x,y\in\mathbb{R}$. We therefore get the following equalities:

$$\forall y,\quad f(1)+g(y)=y,$$

and

$$f(0)+g(y)=0.$$

Hence, it follows that $f(1)-f(0)=y$ for all $y$. But this cannot be because $f(1)-f(0)$ is a constant.

Now, suppose there exist functions $f$ and $g$ satisfying $f(x)g(y)=x+y$. We therefore get the following equalities:

$$\begin{align*} f(1)g(-1)&=0, \\ f(1)g(0)&=1+0=1, \\ f(0)g(-1)&=0-1=-1. \end{align*}$$

From the first equality, we infer that either $f(1)=0$ or $g(-1)=0$. In either case, we stumble into a contradiction. If $f(1)=0$, then it follows from the second equation that $f(1)g(0)=0\ne 1$. Likewise, if $g(-1)=0$, it follows from the third equation that $f(0)g(-1)=0\ne -1$.

b) Constant functions where $f(x)=g(y)=c$ for all $x,y\in\mathbb{R}$ obviously satisfy

$$f(x+y)=g(xy).$$