Assuming a buck converter that's 100% efficient, yes, you will be able to draw 1A from the output.
In practice, buck converters can be very efficient, often over 90%, so you'll still be able to get around 900mA output, which is rather better than the 500mA you'd get with a linear regulator.
An important part of the buck converter system is its input filter ...
As @PeterJennings said, the power dissipated by the polyswitch when it is conducting is a function of current. The polyswitch has its own resistance, which works against the current to generate a voltage, and, hence, heat. The switch then trips when it gets hot.
The 240V part of the rating is the voltage the switch can withstand when it is off. At a ...
The quick answer is that the power is not dissipated in the fuse but in the load. If, for sake of argument, the fuse has an internal resistance of 0.1 ohm then at it's full rated current it will be dissipating I^2R W = 1.6W and experience a voltage drop of IR = 0.4V. The rest of the voltage, be it 15.6V or 239.6V is across the load it is protecting.
Since DC motors are also generators, the reverse voltage raises proportional with the RPM. So the higher the RPM, the higher the reverse voltage -> less current. Also the ambient temperature is most probably not stable, which leads to fluctuating current measurement results.