1) Show that $(\Bbb Z[\sqrt2]^*, .)$ is infinite.
2) Classify $(\Bbb Z[\sqrt2]^*, .)$, where $\Bbb Z[\sqrt2]^*$ is the group of units of $\Bbb Z[\sqrt2]$
What I have done so far that for $a+b\sqrt2$ its inverse will be $\frac {a-b\sqrt2}{a^2-2b^2}$
Now $\frac {a}{a^2-2b^2} \in \Bbb Z$ & $\frac {b}{a^2-2b^2} \in \Bbb Z$
What conclusion can be drawn from it about the cardinality of $\Bbb Z[\sqrt2]^*$?
One can post the answer also for the 1st part only.
Note that $(\sqrt{2}+1)(\sqrt{2}-1)=1$, hence $\sqrt{2}+1$ is a unit of $\mathbb{Z}[\sqrt{2}]$. Then you have that
$$(\sqrt{2}+1)^2, (\sqrt{2}+1)^3, (\sqrt{2}+1)^4, (\sqrt{2}+1)^5, \dots, (\sqrt{2}+1)^k, \dots$$ are all distinct units in $\mathbb{Z}[\sqrt{2}]$.
Finally, one can characterize units of $\mathbb{Z}[\sqrt{2}]$ saying that $$\mathbb{Z}[\sqrt{2}]^* = \{ a+b\sqrt{2}: a, b \in \mathbb{Z}, a^2-2b^2=\pm1\}$$
EDIT: A proof of this fact follows:
Clearly "$\supseteq$" holds, for what you wrote in your question. Viceversa, to prove "$\subseteq$", let $a+b\sqrt{2}$ be a unit. We want to show that $a^2-2b^2=\pm1$.
Let $c+d\sqrt{2}$ be its inverse.Then $$ 1= (a+b \sqrt{2})(c+d \sqrt{2}) = (ac+2bd) + (ad+bc)\sqrt{2}$$ hence $ac+2bd = 1$ and $ad+bc=0$.
Now $$(a^2-2b^2)(c^2-2d^2)=(a+b \sqrt{2})(c+d \sqrt{2})(a-b \sqrt{2})(c-d \sqrt{2}) =$$
$$ =1\cdot (a-b \sqrt{2})(c-d \sqrt{2}) = (ac+2bd) - (ad+bc)\sqrt{2} = 1+ 0 \sqrt{2} =1$$
hence $a^2-2b^2$ is a unit in $\mathbb{Z}$, i.e. it is $\pm1$