Cesko-Slovacko-Poljsko, 2013

a) Let $s = x^2 - rx$ and $t = x^3 - rx$ be rational. Then
$$x^2 = s + rx,$$
$$x^3 = x^2 \cdot x = (s + rx)x = sx + rx^2 = sx + r(s + rx) = (r^2 + s)x + rs,$$
consequently
$$t = x^3 - rx = ((r^2 + s)x + rs) - rx = (r^2 - r + s)x + rs.$$
if $(r^2 - r + s) \neq 0$ (which is $x^2 - rx + r^2 - r \neq 0$), then
$$x = \frac{t - rs}{r^2 - r + s}$$
is rational.
We conclude that the given statement holds iff the equation
$$x^2 - rx + r^2 - r = 0 \quad (1)$$
has no irrational roots. For the sake of completeness, note that if (1) has a rational root then $s$ and $t$ are rational as well:
$$s = x^2 - rx = r - r^2 \quad \text{and} \quad t = 0 \cdot x + rs = rs = r(r - r^2).$$
If the determinant $D = r(4 - 3r)$ of (1) is less or equal 0, then (1) has no real solutions or a solution $x = \frac{r}{2}$, which is rational. Since $D \le 0 \Leftrightarrow (r \ge \frac{4}{3}$ or $r \le 0)$ the statement under consideration is true.

b) According to a) it suffices to show that $D > 0$ and $\sqrt{D}$ is irrational. We have
$$D = r(4 - 3r) = \frac{p}{q}\left(4 - \frac{3p}{q}\right) = \frac{p(4q - 3p)}{q^2} > 0.$$
But $p \nmid 4q$ thus $p \mid p(4q - 3p)$ and $p^2 \nmid p(4q - 3p)$, that is $p(4q - 3p)$ is not a perfect square, that is $\sqrt{D}$ is irrational.