Think of the voltage drops across each resistor. The left drawing (attached) shows this - 1V across R1, 2V across RV, and 2V across R2. The bottom end of the potentiometer is thus 2V higher than 0V (0V + 2V = 2V), and the top end of the pot is again 2V higher than the bottom end (2V + 2V = 4V)
There are a couple of ways to calculate R1 and R2; let's start with Ohm's law to calculate the current through the resistor chain. It's the same current passing through all three resistors (Kirchhoff's current law). Since we know the voltage across RV, and the resistance of RV:
[latex]I = {\frac {V_V}{R_V}} = {\frac{2V}{10k\Omega}} = 200\mu A[/latex]
We now know the current through R1, and the voltage across it, so we can calculate R1's resistance:
[latex]R_1 = {\frac {V_1}I} = {\frac{1V}{200\mu A}} = 5k\Omega[/latex]
Similarly for R2:
[latex]R_2 = {\frac {V_2}I} = {\frac{2V}{200\mu A}} = 10k\Omega[/latex]
You could calculate them another way - by understanding that the voltage dropped across each resistance is in proportion to the resistance. That is, the ratio of resistances is equal to the ratio of voltages dropped:
[latex]R_1 : R_V : R_2 = V_1 : V_V : V_2 = 1 : 2 : 2[/latex]
This means that RV and R2 have the same resistance, and R1 is half of that resistance. The values are easy to see intuitively.
A third way is to use the potential divider rule to derive equations that can be solved by rearrangement:
[latex]V_1 = \frac{5R_1}{R_1 + R_V + R_2} = 1V[/latex]
[latex]V_V = \frac{5R_V}{R_1 + R_V + R_2} = 2V[/latex]
[latex]V_2 = \frac{5R_2}{R_1 + R_V + R_2} = 2V[/latex]
I'll leave you to figure those out.