As a step in an exercise, I'm trying to find voltages \$V_1\$ and \$V_2\$ below
Could a voltage divider (with more than two resistors) be used here? So that, for example:
$$V_2 = \frac{2\ \mathrm{k\Omega}}{2\ \mathrm{k\Omega}+1\ \mathrm{k\Omega}+2\ \mathrm{k\Omega}}\cdot 4\ \mathrm{V} = 1.6\ \mathrm{V}$$
Yes. It does not matter if there is two, three or any other amount of resistors in series, you know all the resistances and voltages with Ohm's law and can use it as much as needed to find a formula for voltage divider with two or any other amount of resistors.
But V2 will not be 1.6V. Otherwise, you are on the right track.
You can certainly have more than one resistor in a divider, in old vacuum tube circuits it was common to derive a number of different voltages needed by the circuitry by just putting a big multi-resistor divider across the the high voltage supply to tap off the lower voltages needed.
A good way to look at voltage dividers is to think of them as taking a part of a whole.
In your circuit you have a 2k, a 1k, and another 2k for a total of 5k. There is 4 V across the whole thing. Now say you want the voltage from ground to V1, so you're taking 2 parts out of 5 (you can forget the k as long as they're all specified in k) so you just think 'two-fifths of 4 V', a fifth of 4 is 0.8, multiply by 2 and you've got 1.6 V. If you want ground to V2 that's three-fifths of 4 V, or 2.4 V. You could find the voltage across just the 1k, one-fifth of 4 V is 0.8 V. With a bit of practice a lot of times you can do the math in your head, otherwise a simple calculator will suffice.
You can work this backwards to design a divider, say you have a 12 V source and you want 5 V, so you think 'I want 5 parts of 12', then you can just make the total resistance some multiple of 12, say 12\$\times\$100 for 1200\$\Omega\$, and you know the bottom resistor must be 5 parts of that so 5\$\times\$100 = 500\$\Omega\$, then just subtract that from the total to get 700\$\Omega\$ for the top resistor.
