Period for a Rubik's cube repeated manipulation

Answer for part $2$ with suggestion for general idea:

The face facing the observer at the end must be green, so the solution is

$$0\mod 4\,.$$

Consider the piece starting at the bottom right of the green face. At the end of each manipulation, it remains at the bottom right of the face facing the observer, so in order to be in the correct place at the end, the face facing the observer must be green.

Moreover after $4$ manipulations the piece has rotated, so for the piece to be in the correct orientation, we must have performed some multiple of $12$ manipulations - that is the solution is

$$0 \mod 12\,.$$

In general, look at when specific pieces or collections of pieces (say corners) return to their original position. You may also cluster manipulations. Knowing the solution is $0 \mod 12$, my next step would be to observe what happens to another piece after $12$ manipulations.

The top right piece (of the green face) for example is moved to the bottom right of the green face, left alone for to manipulations then moved back up, but rotated, hence after $12$ manipulations it is left unchanged.