For a) just note that equation $x^2-2y^2=-1$ has infinitely many soultions in $\mathbb{N}^2$ and for such pair $(x,y)$ we have $\gcd(x,y)=1$ and $[\sqrt{2}y]=x$ if $y\geq 1$.

Let $\sqrt{2}=0.b_1b_2b_3\cdots _{(2)}$. Then we have infinitely many integers $n$ with $b_n=0$ and infinitely many integers with $b_n=1$ ; otherwise $\sqrt{2} \in \mathbb{Q}$. For $b_n=1$, we have $\gcd \left(2^n, \left[\sqrt{2} \cdot 2^n \right] \right)=1$, and $\gcd \left(2^n, \left[\sqrt{2}\cdot 2^n \right]\right)\ge 2$ otherwise. This proves both (a) and (b).

For (b) : Suppose that $x^{2} -2y^{2} =-1$. Then $( 2x)^{2} +4=2( 2y)^{2}$. So we have $2x=\left\lfloor 2y\sqrt{2}\right\rfloor $. Therefore, $\gcd\left( 2y,\left\lfloor 2y\sqrt{2}\right\rfloor \right) =\gcd( 2y,2x) \geq 2$.