The Chinese Remainder Theorem for Rings.

Yes, your solution appears to be complete and correct.

Note that 1 is not necessarily in $R$ for me, so your discussion for (a) is a little bit mistaken. Here is my thought for (a), you can consider it:

We have $I+J=R$, so there exist $i \in I,j \in J$ such that $i+j=s-r$. Put $p=r+i=s-j$, and we get $p$ is a solution of the system since $p\in r+I$ and $p\in s+J$.

Note: $x\in r+I$ is equivalent to $x\equiv r \pmod I$