How to calculate series resistor to provide a certain voltage to a pin?
As you correctly figured out, using a single series resistor to drop voltage requires knowing the resistance of the input it will be driving.  Not only that, it requires that resistance to be fairly constant.  You generally don't know or can't rely on either of those things.

Somewhere there should be a spec for input resistance of the pin you are trying to drive.  Sometimes that is defined in terms of maximum leakage current instead of a resistance.  In either case, that is usually only a minimum resistance (or maximum leakage current).

You need to add a second resistor to ground to form a voltage divider, with the output impedance of the divider being significantly less than the minimum resistance of the input so that the input resistance doesn't matter.  For example, let's say you have a 0-5 V digital output.  If you put 1 k&Omega; in series with that followed by 2 k&Omega; to ground, the output will be 0-3.33 V unloaded.  Converting this to a Thevenin equivalent voltage source, you have a 3.33 V source with a impedance of 1k&Omega;//2k&Omega; = 667 &Omega;.  As long as the effective resistance of whatever this divider is driving is significantly more (like 10 k&Omega;) than the 667 &Omega;, the voltage will be predictable and about 0 to 3.3 V.