>>103289400
V={1, 2, 3, ..., 50}
A={2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47} <-- 15 primes
PS={4, 9, 16, 25, 36, 49, 64, 81} <-- 9 possible perfect square sums
MT={10, 20, 30, 40, 50, 60, 70, 80, 90, ..., 2350 (50x47)}

1. The smallest possible sum of cards that is a perfect square is 4 (Vedal picks 1 and Anny picks 3, or Vedal picks 2 and Anny 2)
2. Too lazy to do but you just count all ordered pairs from the cartesian product VxA whose sum of elements give perfect squares from PS
3. If it's a multiple of 10, then it's divisible by 5 and 2 (primes), one of which come from Anny. If 2 comes from A, then 5 comes from V and there are 50/5=10 options: ((2,5), (2,10), ..., (2,50)). If 5 comes from A, then 2 comes from V and there are 50/2=25 options: ((2,5), (4,5), (6,5), ..., (50,5)). For the rest of A's primes, they need to be multiplied by a multiple of 10 from V themselves, so (10, 20, 30, 40, 50) times the rest of A's primes, (15-2)*5=65. So in total 10+25+65=100 pairs where Anny wins.