Do bijections like the ones defined in this question exist? 
Do everywhere discontinuous bijections $b: \mathbb R \to \mathbb R$ such that $b(\mathbb Q) \cap \mathbb Q= \emptyset$ exist?


Yes, example: $f(x)=x+\sqrt{2}$.
Edit: for the function to be discontinuous everywhere:
$$
 f(x) = 
  \begin{cases} 
   x+\pi & \text{if } x \in \mathbb{Q} \\
   x+\pi+1 & \text{if } x \in \mathbb{R} \backslash  \mathbb{Q}
  \end{cases}
$$
