Here's a thought in another direction, Do you know how much current your port can sink?
If a pull up resistor is supplied from the the PC PSU +5V or a wall wart etc.... to Pin 1 of the coupler, and the port connected to pin 2 the opto-coupler would protect the output from reverse voltage and can likely sink more current than it can source...
A quick look at "a" tlp521 data sheet says
INPUT DIODE
Forward Current 50mA
Reverse Voltage 6V
so 35mA isn't enough for the one I looked up
1.15 typical forward voltage drop from 5 volts = 3.85V/.050 = 77Ohms so thy a 82Ohm 10% or a 75 ohm 5%
at a port high forward V of 5-3.5-1.15 =0.35V/75=4.6mA so you should be good to go.
Same "MIGHT" work with a couple of diodes and a resistor, but no guarantees on this one.... .7v for a regular diode... shotkeys are lower .3ish
supply +5V------>|--------^^^^-----------X---------->|--------+3.5V port
1Kohm
Take your signal at "X" it will boost it .7V to 4.2Vhigh and .7v low... which "should" satisfy CMOS and maybe TTL Levels and only 5mA to sink but the i think the inputs to the drives should be happy with the voltages. (The reason I suggest two diodes is that 1.4V+3.5V=4.9V and 1KOkms on .1v is 100uA so it shouldn't affect the port)
If someone is willing to try this maybe on a PCI card that puts out 3.3 - 3.5 volts Id love to hear the results, because I just thought of this.
My answer for one of your questions is that if, as I believe, an active high means that the port is low until it sends a high step pulse, of a few micro seconds, the low will not supply current so the transistor is OFF and the LED is forward biased with voltage=ON. If the active low puts the port high, the transistor is now biased and is sinking the current, so the LED is OFF. If you go +5V to resistor to LED to transistor the LED will be on when the port is high not low.
I think I spent way to long on this, but as long as this helps you or someone down the road.... all is not lost :p