The A-2 has a min CTR of 0.6, so if you can get 5mA into the LED, you will be able to sink about 3mA at the collector.
I am guessing that your PIC can source 5mA at a voltage drop of ~0.3V, so R1 (designator in your sim) will have 3.0V at one end, and 1.25V (the Vf of the input LED @ 5mA) at the other end, so R1= (3-1.25)/5m = 350Ω.
R2 will have 3.3V at one end, and Vce(sat) at the other when the collector current is 3mA, so R2 = (3.3 - 0.5)/3m = ~ 1KΩ.
Get rid of R3, it only makes it worse....
The speed of the opto is more than sufficient at 9600 baud at these current levels...
Since the configuration you showed originally in your sim inverts, I think you want to wire the input LED so that its anode is tied to 3.3V, and its cathode goes to the transmitting port pin. That doesn't change the math enough to worry about.