Let n, m be the number of disks and the number of piles at a moment. Each movement will rduce n by one, but increase m by one. So n+m = 1000 + 1 = 1001 Since there are no empty piles, when only 3-disk piles exist: 3m + m = 1001 4m = 1001 Which is contradictory. Similarly, since 2002 is not a multiple of 4, point b) is solved.

The same solution applies to point B (edited in the above.)

Another solution is possible.We prove that all $n$ for which it is possible to make all piles of size $3$ are of the form $n=4k+3$.The base case is $k=0$.Also,we have for our base that it is not possible for $n=0,1,2,4$.Now,we prove this claim by induction.Suppose that it is possible for some pile.Let its size be $n$.Denote the size of $2$ piles in which it is separated by $p,q$.We have by our base that $p$ and $q$ are of the form $4k+3$,so $s=p+q+1$ is also of the form $4k+3$,so we are done.