What opto boards are you using?

So a diode is pretty much helping you do a voltage divider circuit. Since they give you the current, and you know the breakdown (forward voltage), the formula is:

(Vsupply - Vforward)/current = series resistor.

With a 12 volt supply:
(12-1.7)/0.100 amps = 103 ohms <<-- this is the smallest size resistor to use

(12-1.7)/0.075 amps = 137 ohms for recommended range.

I would do a 120 ohm series resistor from your power supply.

Oh the wattage: p=v x I
So you will want to use at least a 1 watt resistor.

Of course a smaller power supply the better. Like maybe drive from a 5 volt source.

How do you know if the transmitters are working when the receivers are not working? They are used as a pair. The only way I can see is if the output of the transmitter is visible to a digital camera that does not have an infrared filter. If you provide circuit diagrams this could be more easily verified.

If you use an appropriate current limiting resistor you should not damage the transmitter. If you haven't used a current limiting resistor and have applied power to the transmitter you may have damaged it. Having said that I'm going to assume you have a valid transmitter side circuit and would agree with you that the transmitter is probably working fine. The transmitter side circuit is somewhat simple. Note that Williams drove their transmitters around 40mA so I think 75mA is quite high.

I'm going to assume that you don't have a receiver circuit as the reason you asked the question.The 1.7V specified as the forward voltage is for the transmitter not the receiver. The receiver is a photo-transistor not a light emitting diode. You need to design a receiver circuit. It will need to detect the binary state of open or closed. When the light from the transmitter falls onto the receiver it should output voltage equal to the supply voltage (the actual output voltage depends on the amount of current flowing through the receiver and the amount of current depends on the amount of light and the amount of light depends on the amount output by the transmitter). When the light is interrupted the voltage should be close to zero but there is still current flowing. This is the "dark current" as specified in the data sheet.

It's likely the transmitter is QED123 and the receiver is QSD124. You can find those data sheets by searching the web or visiting a general electronics merchant site.

We documented optos here: https://docs.missionpinball.org/en/dev/mechs/switches/optos.html#electronical-details. If you are using MPF this is how to configure such a trough: https://docs.missionpinball.org/en/dev/mechs/troughs/modern_opto.html.