How do you figure that? I've not seen the expression ohm/(load quantity) before..?

If you have identical resistors in parallel, the equivalent resistance is divided by the number of resistors.

So for two identical resistors in parallel the equivalent resistance is half. For three identical resistors it's a third, for four identical resistors it's a quarter and so on. Your LED resistors are in parallel with this one resistor. If you want to halve the current you need to use the equivalent resistance to all these parallel resistors. This means you need:

150 ohm / 39 (identical resistors) = ~ 3.9 ohm

90 ohm / 56 (2 chains of 28 identical resistors, all in parallel) = ~ 1.5 ohm

For the power calculation I used P = I * I * R. Using your value of 10mA per LED I got the following:

Current through 3.9 ohm resistor is 10mA * 39 = 0.39A.

Therefore P = 0.39 * 0.39 * 3.9 = 0.6 watts. I'd get a 1 watt resistor to be safe. Current may actually be higher than 10mA.

e.g. Current is actually 11mA per LED and the power dissipated in the 3.9 ohm resistor is 0.72 watts.

Current through 1.5 ohm resistor is 10mA * 56 = 0.56A.

Therefore P = 0.56 * 0.56 * 1.5 = 0.47 watts. Again I'd get a 1 watt resistor.

If you can't get a 3.9 ohm or 1.5 ohm 1 watt resistor you can make them by combining higher value resistors in parallel, but you may not need 1 watt resistors any more. It will depend on the actual values you can buy. Worst case you could wire 39 150 ohm resistors in parallel to make your 3.9 ohm resistor. Each resistor would only dissipate 0.01 * 0.01 * 150 = 0.015 watts.

Iain