CO2 is eliminated b/c there is no carbon source. O3 is unlikely just because you don't see it much and because O3 would have to break an N-O bond and form two O-O bonds, so keep it in the back of your head as a possibility that is unlikely. It comes down to H2 and NO, and at first glance H2 looks good because you know that HNO3 is a SA and loses its H+ quickly.
From your MCAT preparations you can guess that the Cu strip is being oxidized by HNO3, hence bubbles. Cu half reaction is Cu--> Cu2+ +2e- or something to that end, not worried about stoichiometry. We also know the e- are not going to a cathode because it is just a cu strip in a solution. Something in the solution must be reduced.
As mentioned we have H+ and NO3- as options. Although 2 H+ could react with 2e- to form H2 (defined as 0 volts half reaction) we know that the small amount of H+ in the H2O solution before didn't generate bubbles (1E-7 M H+ is enough to do that, theoretically, plus le chatliers would make it favorable), so the only choice left is NO.
To rationalize this if POE isnt satisfying, you can say the N in NO3 has a +1 formal charge and wants the e- well more that H+ (N+ is more EN than H+) and will be more strongly attracted to the free electrons than H+. It does look complicated though, because why does N break two N-O bonds? they are broken by the addition of an extra electron from the Cu.
