I'm getting all the questions in the text right, so I think my calculations are OK
Just imagine a resistor inside the battery. As you are an expert in Ohm's law:
V+ -- Ri --+-- R1 --+-- R2 ----GND 0.28mA
5.7V 330 | 10k | 10k
| |
5.6V 2.8V
V+ -- Ri --+-- R1 --+-- R2 ----GND 6.48mA
5.7V 330 | 270 | 270
| |
3.5V 1.75V
However the Ri size is not a real constant, just an estimate got from your actual numbers.
( As is the assumed "no load voltage" of 5.7V )
It should be clear that a with an internal resistance of about the same magnitude as the load, your 5.x V battery is far away from an ideal voltage source 