About the addition formula $f(x+y) = f(x)g(y)+f(y)g(x)$ Consider the functional equation
$$f(x+y) = f(x)g(y)+f(y)g(x)$$
valid for all complex $x,y$. The only solutions I know for this equation are $f(x)=0$, $f(x)=Cx$, $f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$.
Question $1)$ Are there any other solutions ?

If we set $x=y$ we can conclude that if there exists a $g$ for a given $f$ then $g$ must be $g(x)=\frac{f(2x)}{2f(x)}$. By using this result, I tried setting $y=2x$ yielding
$$f(x+2x)=f(x)g(2x)+g(x)f(2x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$$
Question $2)$ Are $f(x)=0$,$f(x)=Cx$,$f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$ the only solutions to $f(3x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$? If not, what are the other solutions?

You can take $g=1,$ then your equation reduces to $f(x+y)=f(x)+f(y).$ Without any additional assumption on $f,$ this functional equation has plenty of discontinuous solutions. See here 
Of course, once you add some regularity condition (such as continuity, differentiability, boundedness), you can get much more using what Vadim wrote in his post.
As to your question 2 - the same arguments apply, since this is a consequence of the general equation.
Since I do not know which condition you prefer to add, let me just mention that if $g$ is continuous, then equation $g(x+y)=2g(x)g(y)$ has solution $g(x)=0$ or $g(x)=2^{ax-1}.$ To see that, note that if $g(x_0)=0$ at some point then $g(x+x_0)=0$ everywhere. Alternatively, $g(2x)=2g^2(x)>0$ for all $x$ implies that the function $h(x)=\ln g(x)-1$ satisfies $h(x+y)=h(x)+h(y)$ and $h$ is continuos. So it is linear. Note, that I assumed that $f$ maps $\mathbb{C}$ into $\mathbb{R}$ but the situation does not change much if you switch to functions with complex range.
