3) Let s'\in S'=\mathbb{Q}. The function \phi is onto if we can find x so that \phi(x)=7x+1=s'. But that easy x should be the solution to 7x+1=s', i.e., x=\frac{s-1}{7} and it is in Q. Hence \phi is onto

For the whole thing your idea is OK. But to be in order you need to

1) define what u*v means (well you guess this from your calculation in 4)

2) shows \phi(u*v)=\phi(u)+\phi(v) (i.e, \phi is a homomorphism). This calculation guides you to answer 1, but in writing you need to answer 1 first and somehow you you define u*v out of thin air.

3) show that it is injective

4) show it is onto.