AC is enough to guarantee the existence of both Bernstein Sets and Vitali Sets...
However is the existence of Vitali Sets strictly weaker than that of Bernstein Sets?
What about the other way round?
AC is enough to guarantee the existence of both Bernstein Sets and Vitali Sets... However is the existence of Vitali Sets strictly weaker than that of Bernstein Sets? What about the other way round? 


For your second definition of Vitali set, I have a weak partial answer. Namely the existence of a Bernstein set does not imply the existence of a $T$Vitali set. The answer can be found in logic blog maintained by Andre Nies: Added: A The Logic Blog is on the arXiv. According to a comment below, the following is a better reference than the dropbox link:
Note that a Turing degree does not need to be an addition group. I don't know whether the existence of a Vitali set implies the existence of a Bernstein set. But it is not difficult to see, under $ZF+DC$, that there is a Vitali set (if it exists) which contains a perfect subset. For you first definition of Vitali set, I have no idea. 

