Is the quotient $R/(a,b)$ equal to first quotienting $R$ with $(a)$ and then with $(b)$ Is the quotient $R/(a,b)$ equal to first quotienting with $(a)$ and then with $(b)$? 
I've been thinking about this for some time. And I think the following is true:
$$\frac{R}{(a,b)}=\frac{R/(a)}{(a,b)/(a)}\overset{(1)}{=}\frac{R/(a)}{(b)/(a)}=\frac{R/(a)}{bR/aR}\overset{(2)}{=}\frac{R/(a)}{b (R/aR)}=\frac{R/(a)}{[b]}$$
where I define $[b]$ as the ideal generated by $b$ in the ring $R/aR$. So this seems true. For (1), I already asked a question. And for (2) I think it is true, but not yet able to proof it. How can I proof that $b(R/aR)=bR/aR$ ?

What you have is that $(R/(a))/(\overline{b}) \cong R/(a,b)$, where $\overline{b}$ denotes the reduction of $b$ mod $(a)$, i.e. $\overline{b} = b + (a)$ as an element of $R/(a)$.  The isomorphism is induced from the natural map $R / (a) \rightarrow R / (a,b)$ which sends $x + (a)$ to $x + (a,b)$.  You need to check that this is a well-defined homomorphism, that it is surjective, and that the kernel is equal to $(\overline{b})$.  Then apply the Fundamental Theorem of Ring Homomorphisms to obtain the conclusion stated in the first sentence.
