I don't have much to say about (1) or (2), but (3) is, of course, a classical equation and the usual interpretation is this:

I'm assuming that you are talking about a domain in the $z$-plane, also known as the $xy$-plane, and that, by $\Delta f$ you mean the classical $f_{xx} + f_{yy}$, not the negative of this or this times some metric scalar (as some sources interpret $\Delta$). In this case, the general solution of $\Delta f = -\exp(f)$ can locally be written in the form

$$
f = \log\left({8\ h'(z)\overline{h'(z)}}\over{(1+h(z)\overline{h(z)})^2}\right)
$$

where $h$ is a (locally defined) holomorphic function on the domain. In the case that the domain is simply connected, one can take $h$ to be globally defined, but you may have to allow it to be meromorphic.

Geometrically, what is going on is that the conformal metric $ds^2 = \exp(f/2)(dx^2+dy^2)$ has constant curvature $K = 1$ and hence must be constructed by pulling back the standard metric on the $2$-sphere by a conformal map (which one can take to be orientation preserving (in the weak sense), so that it is actually a holomorphic map to the $2$-sphere, thought of as $\mathbb{C}\mathbb{P}^1$ and hence $ds^2$ is the pullback via $h$ of the $K=1$ conformal metric

$$
{{4}\over{(1+w\overline{w})^2}}\ dw\circ d\overline{w}\ .
$$