a) Let's show how Olesya can cut the cake to win. Let $M$ denote the weight of the cake. She cuts it into 3 neighboring pieces weighting $\frac{1}{3}M$ (heavy), and all the rest have zero weight (light). Let's number the pieces clockwise 1; 2; ...; $4n$. Piece number $4n$ is neighboring to 1. Let pieces $4n-3$, $4n-2$ and $4n-1$ be heavy. Andriy can not begin with any of 5 pieces: $4n-4$, ..., $4n$, because otherwise Olesya by her first step will choose 2 heavy pieces and collect in total $\frac{2}{3}M > \frac{1}{2}M$. So Andriy chooses any other piece. Then the one who takes at least one of two pieces $4n-4$ or $4n$ (even if together with one of heavy pieces), loses, because the opponent definitely can take 2 heavy pieces and collect more than half of the cake.

Thus, Andriy and Olesya have to make steps in sector between pieces 1 and $4n-5$. After Andriy's turns $4n+1$ pieces will be taken, and after Olesya's turns $-4n+3$. Thus, after completing the full sector between pieces 1 and $4n-5$ Andriy will make his step and so he will lose.

b) Let's show how Andriy should cut the cake to win. Let's number pieces from 1 to $4n$. Let there be 3 heavy pieces weighting $\frac{1}{3}M$ each, and the rest – light ones of zero weight. Let pieces 1, 3 and 5 be heavy ones. Then Andriy by his first step takes piece 3, then by his second turn he can take one more heavy piece and takes in total $\frac{2}{3}M > \frac{1}{2}M$.

It remains to satisfy the conditions of the task precisely. All the pieces must have different non-zero weight. For this in a) make the small pieces weight, for example, $\frac{i}{5(4n-3)(4n-2)}M$, $i=1; 4n-3$, then their sum
$$\frac{1}{5(4n-3)(4n-2)}M + \frac{2}{5(4n-3)(4n-2)}M + \dots + \frac{4n-3}{5(4n-3)(4n-2)}M = \frac{(4n-3)(4n-2)}{10(4n-3)(4n-2)}M = \frac{1}{10}M.$$
Choose the big pieces weighting $\frac{29}{100}M$, $\frac{30}{100}M$ and $\frac{31}{100}M$. Order of pieces does not play role. The same distribution of weights in b), but with condition, that 3rd piece has weight of $\frac{30}{100}M$.