a) The answer is negative. Suppose it's possible and let $1$ is colored in color "c1". Then all positive integers are colored in the same color. Let $p/q$ be a rational colored in the other color, say "c2". All the numbers $n\frac{p}{q}, n=1,2,\dots $ are also colored in "c2", in particular $q\frac{p}{q}=p$ is also colored in "c2", contradiction. b) It is possible. Let $$A:=\left\{\frac{p}{q} : p,q\in\mathbb{Z}_{>0}, (p,q)=1, \text{and } q \text{ is odd }\right\}$$Then $$B:=\mathbb{Z}_{>0}\setminus A = \left\{\frac{p}{q} : p,q\in\mathbb{Z}_{>0}, (p,q)=1, \text{and }q \text{ is even }\right\}$$Color $A$ with one color and $B$ with the other one.

My solution for a) is same as above. Simpliest construction for b) is : Colour $(0,1)\cap \mathbb Q$ red and colour $[1,\infty)\cap\mathbb Q$ blue.