Units of $\mathbb{Z}[\sqrt {-2}]$, and are $3,5$ irreducible in $\mathbb{Z}[\sqrt {-2}]$?

If there exists a number $n=a+\sqrt{-2}b$ such that $n|3$ and $N(n)\neq 1$ (otherwise the number would be associated to $3$) it satisfies $N(n)|N(3)$ (since the norm is multiplicative). This means $a^2+2b^2|9$ and thus $N(n)=3$. But $a^2+2b^2=3$ means $(a,b)=(\pm1,\pm1)$. This means $\pm1\pm\sqrt{-2}$. A quick verification shows that $3=(1+\sqrt{-2})(1-\sqrt{-2})$. $3$ is reducible

For $5$ a similar reasoning shows $N(n)=5$, and this leads to $a^2+2b^2=5$. Reducing the equation $\equiv 8$ we see that there are no integer solutions (since $x^2\equiv_80,1,4$), and thus $5$ is irreducible. If you were to avoid the "modular" reasoning, a simple observation on $a^2\le 5;2b^2\le 5$ allows us to restrict to the values $a=1,2;b=1$: it is easy to verify by hand that none of these values satisfies the equation

If you have no better ideas, you can almost always just do trial division. The two previous answerers did give better ideas, but I thought you might appreciate a different perspective to help confirm the better ideas.

If $5$ reducible in $\mathbb{Z}[\sqrt{-2}]$, that means there is some non-unit number (a number with norm not $\pm 1$) such that $5$ is divisible by it.

With trial division, we could start almost anywhere. I choose to start with $\sqrt{-2}$. We see that $$\frac{5}{\sqrt{-2}} = -\frac{5 \sqrt{-2}}{2} \not\in \mathbb{Z}[\sqrt{-2}].$$ Then I try $$\frac{5}{2 \sqrt{-2}}, \frac{5}{3 \sqrt{-2}}, \frac{5}{4 \sqrt{-2}}.$$ That last one is roughly $-0.884i$, so that tells me to stop on this line and move to the next line.

$5$ is trivially divisible by $1$, because of course $1$ is a unit. $5$ is not divisible by $1 + \sqrt{-2}$, nor by $1 + 2 \sqrt{-2}$. Long story short, by the time I get to $3 + 2 \sqrt{-2}$, I'm convinced $5$ is irreducible in this ring.

It plays out much differently with $3$. Clearly $3$'s not going to be divisible by $\sqrt{-2}$ any more than $5$ wasn't either. But soon I get to $1 + \sqrt{-2}$ and you how this turns out...

Finding the norm is indeed important for determining the second part of this problem, whether or not $3$ and $5$ are both irreducible.

Note that $N(3) = 9$ and $N(5) = 25$. So, if there exist primes $p$ and $q$ in $\mathbb{Z}[\sqrt{-2}]$ such that $pq = 3$ and primes $r$ and $s$ such that $rs = 5$, they satisfy $N(p) = N(q) = 3$ and $N(r) = N(s) = 5$. We needn't worry about $N(p) = N(q) = -3$ or $N(r) = N(s) = -5$ because $a$ and $b$ are limited to $\mathbb{Z}$ and so $a^2 + 2b^2$ can't be negative.

Plugging $a = \pm 1$ and $b = \pm 1$ into the norm formula $N(a + b \sqrt{-2}) = a^2 + 2b^2$, we obtain $N(\pm 1 \pm 1 \sqrt{-2}) = 1^2 + 2 \times 1^2 = 1 + 2 = 3$. Bingo. This means $3$ is in fact not irreducible, that is to say, it is reducible.

Now, if for whatever reason you choose the "wrong" pair out of the six possibilities, such as $(-1 + \sqrt{-2})(1 + \sqrt{-2}) = -3$, you can bring your result to the "right" side of $0$ by multiplying by one of the units you have already discovered.

We have a rather different situation trying to solve $N(r) = 5$. There doesn't seem to be any solutions. We have $a^2 + 2b^2 = 5$. Subtracting $a^2$ from both sides gives us $2b^2 = 5 - a^2$. Then, halving both sides, we get $$b^2 = \frac{5 - a^2}{2}.$$ Obviously this has no solution in integers. Therefore $5$ is actually irreducible.