First, condense every resistor in the network into a single resistor. This will enable you to find the total current flowing into R5 and hence the entire network.
1. Start with combining R8 and RL. Do that with the folllowing formula: R8Ltotal = (R8*RL)/(R8+RL).
2. Then you can combine R8Ltotal with R7 and R9 just by adding them together. This will leave you with R1 in series with 2 parallel resistors: R8, RL, R7 and R9 all combined into a single resistor - in parallel with R6.
3. Use the same formula as step 1 to condense these two parallel resistors into a single resistor.
4. Which leaves you with 2 resistors in series. Add them together to get the whole resistance of the circuit. Then use ohm's law to find the total current flowing into the circuit. This will also be the same as the current flowing through R5 - according to Krichoff's current law.
5. You then have the first junction the current is flowing into. Recall Kirchoff's current law and the "current divider" equation (similar looking but different to the voltage divider equation - be careful with that!) and use the total resistances seen "looking into" each branch that you previously worked out to find the currents flowing into them. Remember that the current flowing through each series connected resistor is the same.
6. Once you find the current flowing through each one, use Ohm's law to calculate the voltages across them. You can find the power dissipated by a resistor if you know the current through it or the voltage across it.