Start off by considering the resistor at the top. You know the voltage across it (9V) and the resistance, so V=IR tells you that I(1)=5mA
You immediately have that the total current going through the bottom part (split across the two paths) is the same (from conservation of charge), so I(2)=5mA-4mA=1mA
You should also be able to see that the voltage across the bottom bit is 3V, since the potential difference across the whole thing has to add up to 12V and the top resistor has 9V across it. The potential is the same at the base (nominally 0V) and at the bit in the middle (9V), so it's 3V across both of the bottom resistors.
Then R1=V/I=3/0.004=750 ohm. The same thing goes for R2, giving 3k ohm.