a) $p+q^2|p^2+q^2p$ so $p+q^2|(p^2+q^2p)-(p^2+q) = q(pq-1)$. Let $d$ divides $q$ and $p+q^2$. Then $d$ divides $q^2$ and so it divides $p$. But now $d$ divides $p$ and $q$ so $d=1$ So $q$ and $p+q^2$ are coprime we are done. b) Since $p+121|p^2-121^2$ we have $p+121|121^2+11 = 11(11^3+1)= 11\cdot 12\cdot 111$. Obviously $p=11$ works. Say $p\ne 11$ then $p+121| 12\cdot 111$. Then $p+121\in \{1,2,3,4,6,9,12,18,36,37,74,111,148,222,333,444,666,1332\}$ So $p\in \{101,323,1211\}$.

@above same solution, but $323=17 \times 19$ and $1211=7 \times 173$ are not prime

I have done it in this way