How to calculate the area of a surface?

As for Q1, I don't hink there is a very simple parameterization of the whole surface, but - similar to what you did - the surface can be partitioned into 12 or 24 surfaces, which are just a rotated and/or mirrored copies of one another and have a simple parameterization. Which also means that for the computation of the surface area one really only needs a parametrization of one of those base parts, ideally one that is suitable for the computations.

Your parameterization looks interesting, but is probably not so well suited as base for the surface calculation, because the z-coordinate uses "abs" (and is generally a bit complex), so differentiation might prove difficult. I used a somewhat different parameterization that is nicer for computation purposes. For the vizualization I split the whole surface into only 12 parts (see attachment), although for the computation I used only half such a part (which actually matches a base part of your surface partition), because it allowed for more simplification of the integral (at least when doing manually, but Geogebra might not have cared actually).

As for Q2, the surface area is "just" the integral of 1 over the surface, which can be expressed as a double integral where the integrand is the length of the cross product of the partial derivatives of the parameterization (see https://en.wikipedia.org/wi... for a few more details). The computation is actually not even that hard to do manually with the parameterization I used. (The intermediate terms look somewhat complicated at first but can always be simplified nicely, and the integral isn't that hard afterwards.) It actually took me more time to get the computation done in Geogebra. Maybe it exists, but I didn't find a predefined command that would calculate (numerically or otherwise) such a surface integral, so did all the steps separately in the CAS. Took me some time to get Geogebra to find the proper simplications for the integrand. Didn't manage to get Geogebra to find the "simple" form I got manually, but even though the one from Geogebra looked extremely ugly, it was enough that Geogebra could compute the integral (exactly, not just numerically), see attachment.