(i) Let $R$ be the radius of the sphere. Then the area of the $k$-th disk exposed is $\alpha_k = \pi R^2 \left (1 - \left ( 1 - \frac {2k} {n}\right )^2\right ) = 4\pi R^2 \frac {k(n-k)} {n^2}$. We are told that $\alpha_1$ and $\alpha_2$ are integer. But $d = \gcd(n-1, 2(n-2)) \mid 2$, and clearly also $d\mid k(n-k)$ for all $k$. By Bézout's relation there exist integers $u,v$ such that $u(n-1) + v(2n-4) = d$, and then, denoting $k(n-k) = md$, we have $mu(n-1) + mv(2n-4) = md = k(n-k)$, hence $mu\alpha_1 + mv\alpha_2 = \alpha_k$ is an integer. (ii) The area of the sphere is $A = 4\pi R^2$. We have $2\alpha_1 - \alpha_2 = \dfrac {2} {n^2} A$ being an integer $N$, so $A = \dfrac {n^2} {2} N$, while $\alpha_2 = (n-2) N$. Clearly $A$ being integer is equivalent to $(n\equiv 0 \pmod{2}) \lor (N\equiv 0 \pmod{2})$, which in turn is equivalent to $\alpha_2$ being even.