I'm not sure that I understand you. Think of a wire as a pipe - the current is the water flow rate and the voltage (drop) is the pressure drop across the pipe. You can have more flow (current), but it'll require a larger pressure drop to push it through.
The internal and external resistors both dissipate power, in the form of heat. It's just that we define that taken by the external resistor as useful.
The total resistance is 3.3 Ohms, from which you can calculate the current that flows (6/3.3 A). I'd then use
P=I2R to calculate the power converted by the external resistor. You could also calculate the voltage drop across the external resistor and then use
P=IV, where V is the voltage across the external resistor, i.e.
6−0.4I.
It's the power loss when the V is across a component that isn't 'useful'. It's the power supplied when the power does something that we want.
In your power station question, the power dissipated by the supply lines isn't useful for the homes, so is said to be lost. If the power dissipated in the home is put through a resistance to generate heat, we class it as useful, but it's still just electricity being converted to heat - it's just a matter of where.