The anode is where oxidation takes place, the cathode is where reduction takes place.
In the electrolysis of NaBr,
the oxidation is 2Br^- ---> Br2(l) + 2e^-; E = -1.066 volts (that's the anode)
the reduction is 2H2O + 2^e− ---> H2(g) + 2OH−; -0.8277 volts (that's the cathode)
Sodium metal would not form it water, as it would immediately re-oxidize. Also, if the voltage is controlled, oxygen will not form at the anode.
Hope that helped!