(a) Because $x^2-rx$ is rational, we know that $x$ is the root of a quadratic polynomial in $\mathbb{Q}[x]$, hence $x=a+\sqrt b$ for some rational $a,b$ where $b\ge 0$. If $\sqrt b$ is rational, we're done, so assume $\sqrt b\notin\mathbb{Q}$. Then $x^2-rx=(a^2+b-ra)+(2a-r)\sqrt b$ and $x^3-rx=a^3+3ab+(3a^2+b)\sqrt b-ra-r\sqrt b=(a^3+3ab-ra)+(3a^2+b-r)\sqrt b$. These are rational iff $2a-r=0$ and $3a^2+b-r=0$. But then $0\le b=r-3a^2=r-\frac34 r^2=r\left(1-\frac34 r\right)$, which is a contradiction for $r\le 0$ or $r\ge \frac 43$.