This is a very good way to present your question.

I'd be inclined to assume that the two curves come together at 0 volts and 3 volts.

This would be consistent with the three sections of the questions.

Topsquark is correct to say use Ohms law, but I would do it differently, since comparing slopes can be misleading.

So for any V between 0 and 3 volts (ie 0 > V > 3) The current in the bulb is greater than the current in the resistor or Ib > Ir.

So for any of these values Δv is the same so if we take a vertical line upwards on the plot from any ΔV value we meet the resistor curve before the bulb curve.

But ΔV is the same for both bulb and resistor at this value so ΔV/Ir > ΔV/Ib ; where Ir and Ib are the resistor and bulb currents respectively.

The resistor resistance will alway be greater than the bulb resistance in this Δvoltage range.

For example

looking at the curve at ΔV = 2 volts, Ib = 125mA and Ir = 100mA

So R(bulb) = 2/0.125 = 16 ohms

and R(resistor) = 2/0.1 = 20 ohms

At ΔV = 3 volts I make the resistances equal.

That's two parts, what do you think about the third part?