*See note at end!
In addition to the ions from AgF and NaBr, we must also consider the hydrogen and oxygen in water (I'm assuming this is aqueous solution).
If the solution is concentrated, we would expect to see only Ag at the cathode, and only Br2 at the anode. This can be seen from the reduction potentials of the species involved. Br- is easier to oxidize than either O2- or F-, so it gets oxidized first. Ag+ is easier to reduce than either H+ or Na+, so Ag metal will come out first.
If the solution is more dilute (for example, after running for a while), the production of oxygen and hydrogen will become more favorable (this favorability depends upon the concentrations of the ions - see the link for the Nernst equation), and you will start to get some O2 at the anode and some H2 at the cathode (in addition to the Br2 and Ag).
F2 and Na will never be produced. F2 is such a powerful oxidizing agent that it will actually react with water to form HF and O2, so if any did happen to form, it would be consumed. Na, as you probably know, reacts violently with water, so if any Na metal formed, it would immediately be consumed by the water.
NOTE: Actually, it just occurred to me: this might be a trick question. While AgF and NaBr are soluble, AgBr is not, and will precipitate if you actually try to form this solution. The only ions remaining (if you use equimolar amounts of NaBr and AgF) will be Na+ and F-, plus those from the water. Because F2 and Na will not be formed, we will only see O2 at the anode and H2 at the cathode.