You have to start with a closed circuit, so that there can flow current. Do you have the resistor connected between the + and the - of the power supply? Then the voltage difference is 3.3 V. And you use Ohm to calculate the current.
The LED. Did you place that in series with the resistor? Which is how a LED circuit is built: the resistor makes sure that there's not too much current through the LED. Always use one.
If the LED's voltage would be 3 V then the difference between your supply voltage and the LED's voltage would be across the resistor. Kirchhoff is to blame for that. Kirchhoff's Voltage Law (KVL) says that the total of the voltages in a closed loop is zero. So we'll have 1.1 mA through LED and resistor. The 3 V was specified at 20 mA, so we're an end below that. It's normal for a LED to have a lower voltage at low currents.
But note that the 1.1 mA was true for 3 V LED voltage. We're apparently at 2.4 V, so the difference is now 0.9 V, and the current 3.3 mA. If you decrease the resistor value so that the current increases, you'll notice that the LED's voltage will increase as well.
How do you calculate the value? (Here we go again)
\$ R = \dfrac{\Delta V}{I} = \dfrac{3.3 V - 3 V}{20 mA} = 15 \Omega \$
edit re your update of the question
Welcome to the real, imperfect world. What you have at hand is measurement error. This is an important issue in engineering, and handling it properly can be a painstaking process.
You're giving your numbers in three significant digits, that's probably what the multimeter gives you. A multimeter's precision is most of the time expressed as a percentage (relative error) + a "count" (absolute error). A hobby quality meter may for instance have 2 % precision +/- 1 count. The 2 % should be clear: a 100 V reading may actually represent anything between 98 V and 102 V. The 1 count is an error in the last digit. A 5 may actually be a 4 or 6. That's an absolute error and doesn't depend on the value the meter gives you. If you measure 100 V then 1 count represents 1 %, if you read 900 V (same number of digits!) then 1 count is 0.11 %.
Let's presume you have a decent multimeter with 1 % +/- 1 digit precision. Then worst case your values may become
3.28 V - 1 count = 3.27 V, - 1 % = 3.237 V
267 Ω + 1 count = 268 Ω, + 1 % = 270.7 Ω
11.7 mA + 1 count = 11.8 mA, +1 % = 11.92 mA.
3.237 V / 270.7 Ω = 11.96 mA, which agrees well with the 11.92 mA we calculated for worst case. If your multimeter has a 1.5 % precision the calculated current will fall perfectly within the measured value's error range.
This only answers one part of your question, but I think it might be the main place where you are misunderstanding.
How can the Ohm Law be applied in this case?
"Ohm's Law" isn't a law for all circuits. It's a description of one particular type of component: the ideal linear resistor.
It doesn't matter how your power supplies are arranged in your circuit. Ohm's law has nothing to do with power supplies.
The general laws you want to consider that describe how currents and voltages relate in any circuit are Kirchoff's Current Law and Kirchoff's Voltage Laws. These laws, along with specific current-voltage relationships for each type of device (one of which is Ohm's Law), are the main tools for analyzing circuits.
Best Answer
The 1.875 Ohm 'resistance' you calculated is based on the 800mA and 1.5V, so this wouldn't be a resistor you add, but is the equivalent resistance of whatever load if drawing that 800mA (which may not be a resistor). As @Tony Stewart mentioned, the load may not be linear, so e.g. it may draw more current at lower voltage, unlike a resistor which will continue following the V=IR formula as the voltage changes.
If you wanted to step down the 5V source to the 1.5V load voltage, you could use a series dropping resistor, calculated as @Andy aka showed. So you would have this:
simulate this circuit – Schematic created using CircuitLab
R2 would be your 'load', and R1 is used to create the required voltage drop.
(Please note that this is OK for educational purposes, but in a practical circuit you would not use a resistor to drop the voltage is it would be very inefficient.)
Many people use the water analogy to think of electrical concepts as it tends to be easier to visualize. In this particular case you would think of a water source at a high pressure (5V), and requiring a specific flow and pressure (800mA at 1.5V). If you were to connect the high pressure source directly to the load, instead of 800mA you would get 5 V / 1.875 Ohm = 2.67 A. But if you put a long narrow hose in between the source and output, this would restrict flow and cause a pressure (voltage) drop.
I hope that analogy helps you think about Ohm's law.