Naturally, we can't be sure without some extra experimentation, but, personally, I think you've already alluded to the answer. Additionally, your data certainly seems to be trending towards a quite reasonable explanation. Your data makes perfect sense, in my opinion.
I'm going to try "guiding" you towards a few thoughts without just stating my thoughts explicitly, since this is your lab assignment, after all ; )
You mentioned that a nitrous/acetylene flame has a greater thermal energy than an air/acetylene flame. This is ABSOLUTELY correct. You also mention that the purpose of the KCl and lanthanum salt and EDTA, etc. are to act as "releasing agents," which are designed to prevent formation of stable complexes with calcium. Obviously, these stable complexes are undesirable, since they don't always decompose in a flame to yield calcium in its ground state. Our FAA/ES can NOT see calcium unless it is in the ground state when we try to measure it. Would a nitrous/acetylene flame more efficiently decompose these stable complexes compared to an air/acetylene flame? One well-documented FAA/ES interference for caclium is phosphate. (Source: http://www.files.chem.vt.edu/chem-ed/spec/atomic/interfer.html
). Notice that BOTH of your blanks have phosphate, and remember that even DI water will have at least SOME calcium in it. Additionally, your added phosphate, KCl, and EDTA will, certainly, have SOME calcium in them, as well. If this phosphate formed complexes with calcium in your blank, what would happen if the air/acetylene flame did not have enough energy to decompose these complexes? What sort of signal would we measure? If your nitrous/acetylene flame has MORE thermal energy, would it be able to decompose some or ALL of the calcium-phosphate complexes? And, would that INCREASE or DECREASE your signal?
Also, notice that blank B had MORE phosphate in it than blank A, and remember that phosphate is a known interference for calcium. And your signal ended up decreasing. If we had more phosphate in this blank, what would happen to the calcium, and would we expect the signal to increase or decrease?
Lastly, I have one final suggestion. Check the slopes for each calibration curve--for the calibration curve using the air/acetylene flame and for the calibration curve using the nitrous/acetylene flame. Is the slope for the nitrous/acetylene curve larger? If so, by how much? The slope of a calibration curve is a measure of the sensitivity of a technique. The slope tells us how much change in signal we got for a given change in concentration. A really large slope means our method is very sensitive, because for a very small change in concentration, we get a very large change in signal, which means that it is easy to tell even small differences in concentrations apart from each other. The sensitivity, in this example, would help us to evaluate if the nitrous/acetylene flame was more effective at decomposing interferences that remained present in the air/acetylene flame.
I hope this helps. Everyone, please feel free to correct me, if you think I veered off into the wrong direction.