Is there a better way to determine the units in an integral domain?
It really depends on the integral domain, very much. Here, your method is reasonable enough; alternatively, you can use the fact that these are complex numbers; the (complex) norm of the inverse of $z$ is of course $\frac{1}{|z|}$. Computing the complex norm of a nonzero $a+b\sqrt{-2}$ will show you that it is always at least $1$, and is strictly larger than $1$ if $b\gt 0$; this tells you that the only units are in $\mathbb{Z}$, and so must be $1$ or $-1$.
*I read that the units of $R[\sqrt{d}]$ with $d$ square free were determined by $\mathrm{Norm}(\epsilon)=\pm 1$. Is this general? Is this for any square free $d$?
Not as general as you state. For one thing, it would suffice for $\mathrm{norm}(\epsilon)$ to be a unit. But if $R=\mathbb{Z}$, then yes: the norm map amounts to multiplying $a+b\sqrt{d}$ by $a-b\sqrt{d}$. If this is equal to $1$ or to $-1$, then this proves that it is a unit (with inverse either $a-b\sqrt{d}$ or $-a+b\sqrt{d}$). Conversely, if $\epsilon$ is a unit, then there is a $\delta$ such that $\epsilon\delta=1$, and then $1=\mathrm{Norm}(\epsilon\delta)=\mathrm{Norm}(\epsilon)\mathrm{Norm}(\delta)$. This tells you that $\mathrm{Norm}(\epsilon)$ must be a unit in $\mathbb{Z}$, and the only units in $\mathbb{Z}$ are $1$ and $-1$.
The fact that $d$ is square free is important: consider $\mathbb{Z}[\sqrt{-8}]$. This is different from $\mathbb{Z}{\sqrt{-2}}$, because it consists only of those elements of the form $a+b\sqrt{-2}$ where $b$ is even; that is, it is strictly contained in $\mathbb{Z}[\sqrt{-2}]$. So you can run into issues if your $d$ is not squarefree.
Irreducibles: What you give is the definition of irreducible. And no, it is false that every nonzero element is irreducible: for example, $4$ is not irreducible, because $4=2\times 2$ and $2$ is not a unit.
Note that it is not enough to know the image of the norm map; it could be, in principle, that you have two elements with the same norm, one irreducible and one not: having no proper divisor of the norm is necessary, but may not be sufficient for irreducibility.
Showing that it is a Euclidean domain can be done geometrically. There's a nice argument given by Klein; you can see it sketched (for $\mathbb{Z}[\zeta_3]$, where $\zeta_3$ is a primitive cubic root of unity) here.
(3) Will follow from (1): you will find that the primes that remain irreducible are precisely the ones that cannot be expressed as a $x^2+2y^2$ with $x$ and $y$ integers. The fact that other primes cannot be so expressed is actually easy if you consider $x^2+2y^2$ modulo $8$.
Define the norm on $\mathbb Z[\sqrt 3]$ to be $N(a + b \sqrt 3) = \vert a^2 - 3 b^2 \vert$.
Let $\alpha, \beta \in \mathbb Z[\sqrt 3]$ with $\beta \neq 0$.
Say $\alpha = a + b \sqrt 3$ and $\beta = c + d \sqrt 3$.
Notice that
\begin{align*}
\frac\alpha\beta &= \frac{a + b \sqrt 3}{c + d \sqrt 3} \cdot \frac{c - d \sqrt 3}{c - d \sqrt 3} \\
&= \frac{ac - 3bd}{c^2 - 3d^2} + \frac{-ad + bc}{c^2 - 3d^2} \sqrt 3 \\
&= r + s\sqrt 3
\end{align*}
where $r = \displaystyle \frac{ac - 3bd}{c^2 - 3d^2}$ and $s = \displaystyle \frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $\vert r - p \vert \leq 1/2$ and $\vert s - q \vert \leq 1/2$.
We want to show that $\alpha = (p + q\sqrt 3) \beta + \gamma$ for some $\gamma \in \mathbb Z[\sqrt 3]$ such that either $\gamma = 0$ or $N(\gamma) < N(\beta)$. (We'll show the latter holds always.)
Define $\theta := (r - p) + (s - q)\sqrt 3$ and define $\gamma = \beta \cdot \theta \in \mathbb Z[\sqrt 3]$ and observe that
\begin{align*}
\gamma &= \beta \cdot \theta\\
&= \beta ( (r - p) + (s - q)\sqrt 3)\\
&= \beta (r + s\sqrt 3) - \beta(p + q\sqrt 3) \\
&= \beta \cdot\frac\alpha\beta - \beta (p + q\sqrt 3) \\
&= \alpha - \beta (p + q\sqrt 3)
\end{align*}
Hence we have $\alpha = \beta(p + q\sqrt 3) + \gamma$.
Finally notice that
\begin{align*}
N(\gamma) &= N(\beta \cdot \theta) \\
&= N(\beta) \cdot N(\theta) \\
&= N(\beta) \cdot \vert (r - p)^2 - 3 (s - q)^2 \vert \\
&\leq N(\beta) \cdot \max\{ (r - p)^2, 3(s - q)^2\} \\
& \leq\frac34 N(\beta)\\
&< N(\beta)
\end{align*}
The key here was that $\vert (r - p)^2 - 3 (s - q)^2 \vert \leq \max\{ (r - p)^2, 3(s - q)^2\}$ since $(r - p)^2, 3(s - q)^2 \geq 0$ and then we use that $(r - p)^2 \leq 1/4$ and $3(s - q)^2 \leq 3/4$.
Best Answer
For $n$ negative this is relatively easy - the only such $n$ are $-1,-2$. For other negative $n$, this ring is not a UFD: $2$ is irreducible (since there are no elements of norm $2$ in $\mathbb Z[\sqrt{n}]$) and divides either $\sqrt{n}\cdot\sqrt{n}$ or $(1+\sqrt{n})(1-\sqrt{n})$, but doesn't divide either factor. Note that for $n\equiv 1\pmod 4$ this is sometimes remedied by working in the full ring of integers of the quadratic field (which is $\mathbb Z[\frac{1+\sqrt{n}}{2}]$ for $n\equiv 1\pmod 4$ squarefree), which is a UFD or Euclidean for a handful more values of $n$.
For positive $n$, the situation is very complicated. It is conjectured that then $\mathbb Z[\sqrt{n}]$ is a Euclidean domain iff it is a UFD, and when the latter holds is somewhat more understood, so let me focus on that.
Firstly, in order for it to be a UFD, $\mathbb Z[\sqrt{n}]$ has to be the full ring of integers in $\mathbb Q(\sqrt{n})$. Unless $n$ is a square, this forces $n$ to be squarefree and congruent to either $2$ or $3\bmod 4$. Further, we can restrict the number of prime factors $n$ can have - it follows from Gauss's genus theory that the $2$-torsion in the narrow class group has order $2^{\omega(n)-1}$, and hence that of the class group is at least $2^{\omega(n)-2}$. Furthermore, if $n\equiv 3\pmod 4$ then the two are equal (since $a^2-nb^2=-1$ has no solution). Since for these rings being a UFD is equivalent to having trivial class group, this means $\omega(n)\leq 2$ if $n\equiv 2\pmod 4$ and $\omega(n)=1$ otherwise. Hence $n$ is either a prime $3\pmod 4$, or twice an odd prime. Both of these cases occur, as indicated for instance by $n=3,6$. Those specific $n$ are also known to give (norm-)Euclidean domains. If we were to consider full rings of integers, then some products of two odd primes would also appear, but that's it.
Unfortunately, this is about where our knowledge ends. It is conjectured that there are infinitely many values of $n$ for which these rings are UFDs and EDs, and in fact it is expected that this holds for a majority of prime values of $n$, but this is far from known. You can find some list of known values on OEIS (ignoring cases $n\equiv 1\pmod 4$, as then your ring is not the ring of integers).
As for the least unknown cases - this is the sort of problem where (with current knowledge, thanks to Harper's work) the problem is essentially reduced to a computational problem, so it is hard to pin down what is the least unknown case. Harper himself checked this in his PhD thesis when $4n\leq 500$ (and indeed all UFDs are EDs in this range), but I have no idea if and how far this has been pushed since 2004.
Edit: I have found a paper of Narkiewicz which proves that for real quadratic fields, the "UFD implies ED" holds with at most two exceptions.