Martins answer to the second question refers to "using a voltage source" and "using a current source". You are not using a voltage source, and you are not using a current source: you are using a two-cell alkaline battery.
Given that you've got a 3.08V circuit, I expect that (1) The battery will discharge rapidly and wastefully initially, then (2) The light will drop out while the battery is still almost fully charged.
In between (1) and (2) you should get a brief region where most of the battery discharge is going to power the LEDs (rather than heating the cells or just switched off). If the cell discharge curve exactly matches the LEDs voltage curve, this could be most of the energy in the cells, but I doubt I would be that lucky. At 3.08V with alkaline cells, I would expect the crossover to be fairly short, and up near the fully-charged end of the discharge curve (but it's the kind of thing you could measure)
If you use 3 cells and a resistor, you waste energy in the resistor, but I think you will use far more of the discharge curve of the cells, and not leave them almost fully charged when they dip below the LED voltage. So you may pay less for cells, even though you use three at a time instead of 2.
Obviously, the main problems with the simple circuit are blowing up the LED or exploding the cells, but clearly that hasn't happened with your experiment, which supports my idea that you are operating up near the top voltage of the cells.