I am trying to answer the following question
For (a) I have said that a and ab are in the ring R, by the definition of a ring. Therefore, by the definition of a Euclidean domain a=abq+r. As we are told that r is non-zero, we know that $\delta(r)<\delta(ab)$.
For (b) I am less sure how to prove this. We know that $\delta(r)<\delta(ab)$ and $\delta(ab)\ge\delta(b)$, but I'm not sure how to show what is required.
For a), you are not given that $r \neq 0$, you need to show this. By definition of an Euclidean domain, you can find $q,r \in R$ but either $r=0$ or $\delta(r) < \delta(ab)$. Suppose $r=0$, then $a=abq$ which implies since $a \neq 0$ and you are in an integral domain that $1=bq$, but that means $b$ is invertible (a contradiction). So $r \neq 0$.
For b), we have $r=a-abq=a(1-bq)$ so $\delta(a) \leq \delta(r)$.