Electronic – How to calculate resistance when I have the voltage drop

400 V seems a stretch for that. The LDR is claimed to be able to dissipate 400 mW, which is plausible enough. To dissipate 400 mW at 400 V would mean the resistance is 400 kΩ, which is also possible for a CdS cell like the one shown. However, whether it can really withstand 400 V without breakdown is another matter.

Since the price seems high and there doesn't seem to be a datasheet, I'd move along and forget about this one.

What the 400 V rating is saying (whether it is correct or not is another issue), is that this is the maximum voltage you can ever put accross this device. That does not mean you have to, and in any normal light sensing circuit you would put much much less than that accross it. This device will operate fine (vary it's resistance as a function of light) at any voltage down to 0. You can use this in a 3 V battery-operated circuit, for example.

This is because of what you might call 'differential impedance': the change in current per unit change in voltage. A resistor's differential impedance is linear: for every extra volt across it, a fixed number of additional amps flow through it, determined by that resistor's resistance value.

Semiconductors don't exhibit linear differential impedance; a unit change in voltage won't always create the same change in current. If you plot voltage vs current for a diode, you'll see something like this:

This is what creates the diode's characteristic voltage drop. The curve is sharp enough that it can mostly be approximated by saying it has a constant voltage drop, but you can see that that's not true - the voltage drop will gradually increase with current. This is easier to see on a log scale graph:

The upshot of all of this is that in a circuit with, say, a resistor and diode in series, they will find an equilibrium, where the current and voltage through each part forms a point on their respective IV curves. Since a diode's IV curve is very steep, the voltage across it shifts very little with current, and most of the voltage will be dropped across the resistor instead.