Amplify current and voltage to preserve battery
Consider conservation of energy.

Very roughly, 1.2 V * 50 mAh is about 215 J.

5 V * 500 mA is .25 W. 

At a .25 W rate of use, 215 J will be consumed in about 450 seconds, or 7.5 minutes.

Edit 

Let's look at the chips you mention.

TPS61200 is a boost converter. It will "boost" the power supply voltage from 1.2 V to some higher voltage --- 5 V in your case. But it will draw more current from the battery than it delivers to the load. This is because it is not magical and it is not able to create energy from nothing.

L272 is an op-amp. It can deliver an output current of up to 700 mA, controlled by a lower power input. But it must get the energy to deliver that output from it's power supply pins.  You must provide a +/- 12 V supply to use this amplifier according to its specifications. If have some other source of power to provide the +/- 12 V, then you can certainly use the L272 to control a 5 V / 500 mA signal powered by that supply. But the power will be taken (very inefficiently) from the +/- 12 V supply, not magically created by the L272.