Yes. Consider calling p with n=1. Then it takes one call. Now call it with n=2. It gets called twice with n=1, and you need to do the multiply. Thus T(2)=2

Yes. Consider calling p with n=1. Then it takes one call. Now call it with n=2. It gets called twice with n=1, and you need to do the multiply. Thus T(2)=2 calls + 1 multiply. Now call it with n=4. Time taken is 2T(2) plus a multiply. Hence T(4)=2(2c+m)+m=4c+3m. Inductively we can prove that T(n)=n * c + m, where c is the time of a call, and m is the time of a multiply, and I'm ignoring small details. If you save the result of a call and reuse it then you get a different result, of course, and if you really want to learn how this works then you should try to figure it out.