What do we know about $NP \cap co-NP$?
There is no decision problem that is (unconditionally) known to be in $coNP \setminus NP$.  If we had a decision problem that we could prove is in $coNP$ and could prove is not in $NP$, then we would have proven that $NP \ne coNP$, from which it follows that $P \ne NP$.  In other words, if we knew of such a problem, a proof that is in $coNP \setminus NP$ would provide a proof that $P \ne NP$.  But of course we don't know of any proof that $P \ne NP$; that is a famous open question.

For similar reasons, there is no decision problem that is known to be in $NP \setminus coNP$.

The closest we have is the following: we know of decision problems that are in $NP \setminus coNP$, if $NP \ne coNP$.  For instance, if $NP \ne coNP$, then SAT is in $NP \setminus coNP$.  (You can replace SAT with any other problem that is known to be NP-complete.)  It is widely conjectured that $NP \ne coNP$, so SAT is a good candidate for such a  problem -- but we don't know any proof that $NP \ne coNP$ (that's another famous open problem).

We can also say that if $NP = coNP$, then there is no decision problem in $NP \setminus coNP$.  From this plus the above discussion, we obtain the following useful fact:


  There is a problem in $NP \setminus coNP$ if and only if $NP \ne coNP$.  If there is any problem in $NP \setminus coNP$, then SAT is one such problem.


The same remains valid if you replace SAT with any other NP-complete problem.  Symmetrically, we also know:


  There is a problem in $coNP \setminus NP$ if and only if $NP \ne coNP$.  If there is any problem in $coNP \setminus NP$, then TAUTOLOGY is one such problem.


You can replace TAUTOLOGY with any other problem known to be coNP-complete.

So, this is pretty much a complete characterization of those classes, up to our lack of knowledge whether $NP=coNP$ or not.