Let's provide the winning strategy for Mykhailo. Suppose that before Fedir's move, the piles contained $(2n, 2n+1, 2n+2)$ stones for some integer $n > 1$. Then there are 3 possible moves.

If he takes first two piles, then, as $(2n, 2n+1) = 1$, after his move we will have the following numbers of stones: $(2n-1, 2n, 2n+2)$. After this Mykhailo chooses the last two piles, and as $(2n, 2n+2) = 2$, after his move the numbers will be: $(2n-2, 2n-1, 2n)$. This is the initial configuration.

If Fedir chooses first and the third piles, then, as $(2n, 2n+2) = 2$, after his move we will have the following configuration: $(2n-2, 2n, 2n+1)$. Mykhailo chooses last two piles once again, and gets the configuration $(2n-2, 2n-1, 2n)$ again.

If Fedir chooses the second and the third piles, then from $(2n+1, 2n+2) = 1$, we see that after his move we will have the following configuration: $(2n, 2n, 2n+1)$. In this situation Mykhailo just takes first two equal piles and wins, as after his move there will be two empty piles: $(0, 0, 2n+1)$.

The initial configuration is just $n = 50$. After the moves of Fedir and Mykhailo we will get the configuration for $n = 49$ (or Mykhailo will win). And so on. After $48$ moves, unless Mykhailo has already won, we will have $(4, 5, 6)$. Fedir will change it to one of: $(3, 4, 6)$, $(2, 5, 4)$ or $(4, 4, 5)$. It's easy to see that in all of them Mykhailo wins in the next move.