Questions about LEDs in a circuit
Yes
No, practically speaking this will not happen because the 3V battery is not an ideal voltage source. It has a non-ideal source impedance, sometimes referred to as its internal resistance, which we can model with a resistor. For example, a CR2032 has an internal resistance of around 15 ohms and a forward voltage of around 3V. If you connect a 2V LED to this, you would expect around (3-2)/15 = 66mA of forward current. Depending on the battery chemistry and physical dimensions, this internal resistance could be larger or smaller. You can read more about internal resistance here.
Not really - if you're putting enough current through an LED to significantly change its forward voltage (i.e. from 2V to 3V, a 50% increase) you're almost certainly going to have released its magic smoke.
If the power supply cannot supply the required current, this is due to one of two things. It could be the internal resistance situation described above, where the rest of the voltage is dumped across the non-ideal source resistance of the supply. Alternatively, it could be due to an active current limiting circuit within the supply which dynamically reduces the output voltage of its output in order to maintain a maximum output current. The second regime is actually the basis of constant-current drive circuits which are commonly used to drive current mode devices such as LEDs!