Is Stepping Down More Efficient When Voltages are Already Closer Together?
No, not really, within limits.  All your voltages are low enough so that special "high voltage" techniques and parts don't need to be used.  With only 7.2 V in, you have to think carefully about making sure the switch can be driven to fully on.  That means getting a FET with low gate voltage requirements, or making a higher voltage.

For a buck converter, the best is probably when the output voltage is 1/2 the input voltage.  That minimizes short pulses on either side since the duty cycle will be 50%, at least in continuous mode.

Another part that affects efficiency is the output voltage by itself.  At 5 V out, even the drop across a Schottky diode is significant.  Synchronous rectification is therefore important if you are trying to push efficiency.

All in all, I'd rather have 12 V than 7.2 V as input for a buck switcher that has to make 5 V out.  However, both can be used quite efficiently.  There is no strong preference, especially if you are using a off the shelf buck converter chip.  In that case, just get a chip intended for the voltage range, with the efficiency you want.

As Ali80 points out in his answer, this is assuming synchronous rectification, which many off the shelf chips now include.  If not, the diode losses dominate and the lower the input voltage the better.  See Ali's answer for details.

With synchronous rectification, there will still be some preference to higher or lower input voltage.  Usually this doesn't matter much, which is why I said "not really" in the first sentence.  If this level of detail matters to you, then you need to carefully read the datasheet for whatever part you are using, and compare datasheets carefully to decide what part to use in the first place.