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?

 A: Yes, example: $f(x)=x+\sqrt{2}$.
Edit: for the function to be discontinuous everywhere:
$$
 f(x) = 
  \begin{cases} 
   x+\pi & \text{if } x \in \mathbb{Q} \\
   x+\pi+1 & \text{if } x \in \mathbb{R} \backslash  \mathbb{Q}
  \end{cases}
$$
A: Yes. For example, define $f:\Bbb R\to\Bbb R$ by setting $f(x)=x+1+\surd2$ unless $x$ is rational, when $f(x)=x+\surd2$. It is easy to see that $f$ is everywhere discontinuous since, at any point $x\in\Bbb R$, there are points arbitrarily close to $x$ where the value of $f$ differs from $f(x)$ by more than $1$. Moreover $f$ is injective. To see this, let $x_1,x_2\in\Bbb R$ such that $f(x_1)=f(x_2)=y$. If $y$ is of the form $q+\surd2$, where $q\in\Bbb Q$,  then $x_1=y-\surd2=x_2$, and similarly $x_1=y-1-\surd2=x_2$ if $y$ is not of that form. To see that $f$ is surjective, take any $y\in\Bbb R$ and note that $y=f(y-\surd2)$ if $y$ is of the form $q+\surd2$ with rational $q$, while $y=f(y-1-\surd2)$ otherwise. Thus, since $f$ is both injective and surjective, it is a bijection.
