a) There exist such numbers: take $a$ to be any transcendental number and $b=1-a$ then $a^n+(1-a)^n=q$ for $q\in \mathbb{Q}$ and $n\geq 2$ would mean $a$ is algebraic, contradiction b) answer is no, proof ad absurdum. $-\frac{a^3+b^3}{a+b}+a^2+b^2=ab \not \in \mathbb{Q}$. But we also have $a^{n+2}+b^{n+2}=(a^n+b^n)(a^2+b^2)-a^2b^2(a^{n-2}+b^{n-2})$ and $a^{n+3}+b^{n+3}=(a^n+b^n)(a^3+b^3)-a^3b^3(a^{n-3}+b^{n-3})$ so both $(ab)^2, (ab)^3$ are rational and hence $ab$ is rational, contradiction.