 Solution

Coprimality gives polynomials $a,b\in \mathbb{Q}[x]$ with $ag+bh=1$, so the ideals $(g)$ and $(h)$ are comaximal. AlsoIndeed, if $u=gr=hs$, multiplying $ag+bh=1$ by $u$ writes $u=aghs+bhgr$, a multiple of $gh$; the reverse inclusion is immediate. Part (b), with $f=gh$, givesThe explicit isomorphism takes the class of a polynomial $r$ to its classes modulo $g$ and $h$.