One doesn't need to use specific surfaces to compute these fields. <PERSON>'s Law works for any surface. The only requirement for <PERSON>'s Law to hold is that the interaction caused by the field must be inverse square with respect to position, such as <PERSON>'s Law:
$$\overrightarrow{F}=\frac{q_i\,q_j}{4\pi\epsilon_0\,r^2}\hat{r}$$
(where the unit vector points from the field source and towards the probe charge), and superposition must hold in order to sum or integrate the source contributions.
Note that another field that satisfies these requirements is the gravitational field of Newtonian Mechanics:
$$\overrightarrow{F}=-\frac{GmM}{r^2}\hat{r}$$
Gauss's Law for this field is:
$$\oint_{\partial V}\overrightarrow{E} \cdot \overrightarrow{dA}\, =\, \int_V-4\pi G\rho\,dV $$
As to why certain surfaces with geometric symmetry are chosen for problems- even though any surface geometry is fine- it comes down to vast simplification in achieving the result. For example, for a uniform sphere of charge, note how the field is always radial to its source. While you could enclose the sphere in any type of Gaussian surface, if you enclose it in a sphere, which is geometrically symmetric to to the physical sphere, the dot product on the left-hand side of the equation goes away. 903
This is because the radial symmetry always allows the field vectors at the surface and the normal vectors of the surface area to be parallel; thus resulting in the dot product always simplifying to:
$$\oint_{\partial V}E\, dA\, =\, \int_V\frac{\rho}{\epsilon_0}\,dV $$
If the field vectors and surface norms were not parallel because of symmetry, the flux integral could still be correctly evaluated, but is usually MUCH harder to calculate. There are instances were it may not be solvable analytically. One other benefit that can occur from properly-chosen symmetry is creating a Gaussian surface that has a constant-magnitude field everywhere on its surface; which is clearly the case for the spherical example. Thus, E is now a constant, which leads to:
$$E\oint_{\partial V}\, dA\, =\, \int_V\frac{\rho}{\epsilon_0}\,dV $$
$$E\,\times (Surface Area)\, =\, \int_V\frac{\rho}{\epsilon_0}\,dV $$
$$E\,=\, \frac{1}{\epsilon_0}\,\int_V\frac{\rho}{(Surface Area)}\,dV $$
As for which geometry to choose for a given problem, it will depend on the problem. Try to look for a surface that is radially symmetric and, if possible, yields a constant field strength over the entire surface.
Hope this helped! 903