Does $\{\pi, \mathbb{R}\}$ exist? I don't know the axiomatic construction of set theory, but I'm asking myself a question :
Does the set $\{\pi, \mathbb{R}\}$ exist?
And $\{i, \mathbb{R}\}$?
 A: Yes, they exist. Furthermore, since $\pi \in \mathbb{R}$, we can conclude that $\pi \neq \mathbb{R}$, at least in ZFC. So $\{\pi,\mathbb{R}\}$ has two elements. However, there's no such guarantee that $i \neq \mathbb{R}$. So all we can say is that $\{i,\mathbb{R}\}$ has between 1 and 2 elements. Of course, if we knew how we had defined $i$ and $\mathbb{R}$, we would probably have a definite answer to whether $i = \mathbb{R}$. 
A: One of the basic axioms of set theory is that for any $x,y$ we have that the set $\{x,y\}$ exists, that is, there is a set $z$ whose elements are precisely $x$ and $y$. This holds whether we work in standard set theory (say, $\mathsf{ZFC}$), where everything is a set, or in set theories that allow the existence of "primitive" objects (sometimes called atoms, or urelements). Anyway, if your set theory is reasonable enough to formalize mathematical discourse, it will have an official version of $i,\mathbb R,\pi,$ etc, so certainly $\{i,\mathbb R\}$ and $\{\mathbb R,\pi\}$ are sets.
