Let $r=\sqrt[3]{5}+\sqrt[3]{25}$, then the minimal polynomial of $r$ over $\mathbb{Q}$ is
\[P_{r}(X)=X^{3}-15X+30\]so we simply consider $(mod\;\;P_{r})$ (i.e, quadratic polynomials).
Notice that $r^{2}-10=5\sqrt[3]{5}+\sqrt[3]{25}$, so
\[\begin{cases}
\sqrt[3]{5}=\frac{r^{2}-r-10}{4}\\
\sqrt[3]{25}=\frac{-r^{2}+5r+10}{4}
\end{cases}\]and hence the polynomial $P(x)=16x^{2}+204x-160$ does the job for part a).
While for part b), the RHS is $\frac{13}{2}r^{2}+\frac{2355}{2}r-2729,$ so it can never match the LHS
(because the LHS will always have a form of $kr^{2}+mr+n$ for some integers $k,m,n$, meaning that $r$ would otherwise be a zero of a quadratic rational polynomial!)