Its dependent what will by rated power of the device (current).
What is the voltage drop running through resistor one
This question cannot be answered because you did not specify the current.
If there's nothing else between the ends of the resistor and the power supply, then the voltage across the resistor is 24 volts, and the current through it is 2 amperes.
use Ohm's law: to drop 2 volts, V = I x R 2 = current x resistance resistance = 2 / current. So you need to know the current
To tell you that we need to know the resistance of the entire circuit.
What is the voltage drop running through resistor one
This question cannot be answered because you did not specify the current.
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
You will need to take the resistance of the load into account if you are going to design a voltage divider. The resistance of the load can completely change the voltage ratio of a voltage divider if not factored into the calculation. you can measure or read R(load), then R(needed) = 0.8 R(load)
12 volts
If there's nothing else between the ends of the resistor and the power supply, then the voltage across the resistor is 24 volts, and the current through it is 2 amperes.
Doing it with a single resistor is not a good idea because it can only be done with exactly the right amount of current. If the current is 1 amp, for a voltage drop of 10.8 volts you need 10.8 ohms (volts/amps).
use Ohm's law: to drop 2 volts, V = I x R 2 = current x resistance resistance = 2 / current. So you need to know the current
A resistor in parallel with a voltages source will not cause the voltage to drop, theoretically. To get a 20 volt drop you need a resistance in series, and the number of ohms is 20 divided by the current in amps. If the current is unknown or variable, the voltage can't be dropped by using a resistor.
1.36 volts Ohm's Law: Volts = Amps * Ohms
Don't follow what a '2200watt resistor' is. A resistor spec is measured in ohms. Ohms Law is expressed as: Voltage drop = current x resistance, and the wattage of the resistor is = volts drop x current. You have to decide if your resistor is 2200 ohms, or is taking 2200 watts. These two alternatives will give different results for the current. If it is 2200 watts, at 110 volts, the current is 20 amps. If it is 2200 ohms, at 110 volts, the current will be 50 milliamps. (0.05amps)
The size of the resistor will depend on the load. Let's look at this a bit to see if we can make sense of it. You want to drop the applied voltage to a device from 12 volts AC to 11 volts AC. That means you want to drop 1/12th of the applied voltage (which is 1 volt) across the resistor so that the remaining 11/12ths of the applied voltage (which is 11 volts) will appear across the load. The only way this is possible is if the resistor has 1/11th of the resistance of the load. Here's some simple math. If you have an 11 ohm load and a 1 ohm resistor in series, you'll have 12 ohms total resistance ('cause they add). If 12 volts is applied, the 1 ohm resistor will drop 1 volt, and the 11 ohm load will drop the other 11 volts. A ratio is set up here in this example, and each ohm of resistance will drop a volt (will "feel" a volt) across it. See how that works? If the resistance of the load is 22 ohms and the resistance of the (series) resistor is 2 ohms, each ohm of resistance will drop 1/2 volt, or, if you prefer, each 2 ohms of resistance will drop 1 volt. The same thing will result, and the load will drop 11 volts and the series resistance will drop 1 volt. That's the math, but that's the way things work. You'll need to know something about the load to select a series resistance to drop 1/12th of the applied voltage (which is 1 volt) so your load can have the 11 volts you want it to have. There is one more bit of news, and it isn't good. If your load is a "dynamic" one, that is, if its resistance changes (it uses more or less power over the time that it is "on"), then a simple series resistor won't allow you to provide a constant 11 volts to that load. What is happening is that the effective resistance of the load in changing over time, and your resistor can't "keep up" with the changes. (The resistor, in point of fact, can't change its resistance at all.) You've got your work cut out for you figuring this one out.