If $\alpha$ is a quadratic integer in $\mathbb{Q}[\sqrt{d}]$, then define a notion of congruence $\pmod{\alpha}$. 
If $\alpha$ is a quadratic integer in $\mathbb{Q}[\sqrt{d}]$Q, then define a notion of congruence $\pmod{\alpha}$. Furthermore, define $+$, $−$, and $\times$ for congruence classes, and show that this notion is well-defined.

I believe that the notion of congruence $\pmod{\alpha}$ means 
$x+y\sqrt{d}\equiv z+w\sqrt{d}$,
$x\equiv z$ and $y\equiv w$ modulo $\alpha$. Also, I was able to define $+$ for all congruence classes, which would be $[a,b]+[c,d]=[a+c,b+d]$.
What would the congruence classes be for $-$ and $\times$? Also, how would I be able to show that this notion is well-defined?
 A: Since subtraction is just addition in reverse, e.g., $x - y = x + (-y)$, you can focus your efforts on addition and multiplication.
The next step, I think, is to review what congruence means in $\mathbb Z$, since you're familiar with that domain of numbers, right?
Given plain integers $n$, $r$, $m$, what does it mean for $n \equiv r \pmod m$ to be true? It means that there is some integer $q$ such that $n = qm + r$. But a lot of times we don't really care what $q$ is.
Likewise in a quadratic integer ring. If $n$, $r$, $m$ are integers in a particular ring of $\mathbb Q(\sqrt d)$, $n \equiv r \pmod m$ means that $n = qm + r$, where $q$ is also an integer in $\mathcal O_{\mathbb Q(\sqrt d)}$, but we aren't terribly concerned about it for our purposes.
One of the commenters suggested looking at $\langle 2 + i \rangle$ in $\mathbb Z[i]$ and its cosets. I think $\langle 4 + \sqrt{14} \rangle$ in $\mathbb Z[\sqrt{14}]$ might make a better example for you.
If $a$ and $b$ are "plain" integers such that $(a - b \sqrt{14})(a + b \sqrt{14}) = N$ is an odd integer in $\mathbb Z$ (meaning that $N \equiv 1 \pmod 2$), I assert that $a \pm b \sqrt{14} \equiv 1 \pmod{4 + \sqrt{14}}$.
For example, $(7 - \sqrt{14})(7 + \sqrt{14}) = 35$. We see that $$7 - \sqrt{14} = (23 - 6 \sqrt{14})(4 + \sqrt{14}) + 1$$ and $$7 + \sqrt{14} = (5 - \sqrt{14})(4 + \sqrt{14}) + 1.$$ Looking at specific numbers should really help clarify these easy concepts that have been obscured by jargon.
