2) seems to be harder (unless I'm missing a simple solution). First, the question as stated is wrong, since the result is clearly false when i=0. So we want to prove it for .
The only way I can see to do this is to quote the result that the multiplicative group of nonzero elements of is cyclic. Let r be a generator of this group. Then , because the elements are just 1,2,...,p-1 in some order. But if then the map consisting of multiplication by i is a bijection of . So the terms are also just 1,2,...,p-1 in some order. That reduces the problem to showing that . But that is obviously true because .