That formula isn't quite right (plug in the rat getting it on the first try). It seems you have everything, but the fact that he chooses correct on the current trial. For example, suppose the rat chooses correctly on the 2nd try. By your formula the rat has a .25 chance of that happening, which actually refers to him choosing correctly on the first try. So it should be:
Where the 1/3rd refers to him getting it on the xth trial beyond the first.
The expectation for this would be done as you do any expectation - the sum of of x times the probability of x. However - the actual calculation is probably going to be super tedious. It simplifies down, but you are having to do a kinda crummy summation.
The same goes for your second problem. You know the summation of all those probabilites have to add up to 1; so using the definition you're going to have to use some series trickery to figure out what A is, as well as the expecation.
I'll maybe work on it later if you haven't gotten it down yet. Good luck!