Let's look at the easy one. OCLL has four twisted corners, but the twist of the last one is determined by the other three, so there are 3^3 = 27 cases. Some cases can have rotational symmetry - there is only only one case that has no twisted corners, and that can't be changed to any other case with an AUF. The H case has a different type of symmetry. There are only two H cases - one with the arms pointing to the front and back, and one with the arms pointing left and right. A U2 of an H case doesn't change the case, but a U of U' does. For the remaining cases, there is no rotational symmetry. For example the headlights OCLL has four distinct forms - headlights facing forwards, left, back or right.

So the number of cases is

(6 cases * 4 symmetries) + (1 case * 2 symmetries) + (1 case * 1 symmetry) = 27

If you ignore symmetry, it is 6+1+1 = 8, and ignoring the OCLL skip case, it is 7 cases.

With PLL, there are 72 cases, but skip, H, Na and Nb can't be transformed into other cases by AUFing. Z and E have U2 symmetry so there are two Z and two E perms that can be transformed into each other with a U move. The rest all have no symmetry, so a U move cycles between 4 distinct cases.

That gives (16 * 4) + (2 * 2) + (4 * 1) = 72

But ignoring symmetry, it's (16 + 2 + 4) = 22, or 21 of you exclude the skip case.

This method of listing out all the cases and then counting symmetries works regardless of the restrictions on the cases, so you can still use it to work out the number of cases if CP is solved*

* assuming that by CP solved you actually mean that the corners are such that they can be solved with R and U moves only. If you actually mean all of the corners are literally in their correct positions then there can't be any rotational symmetry which actually makes things easier.