How to drive 2/3 color LED's and circuit schematic
The 3 colour LED package contains 3 separate LEDs, each for one of the 3 colours.
In the device you refer to they share a common Anode (positive) connection on pin 4 and each LED's Cathode (negative) connection ON A SEPARATE PIN.

On the diagram below from the LED's datasheet I have used a red line to show the relationship between the package pin and the schematic for the common Anode (+ve) connection and green lines for the 3 separate Cathode(-ve) connections.

The 2 colour LED contains 2 separate LEDs -  connection as per datasheet.

To drive a 3 colour LED with common Anode you need 3 of the (slightly modified) circuits that you have shown - one per LED. 
The circuit needs to be modified because you have the current controlling resistor in the Anode connection - but as this is common to all 3 LEDs and you need one resistor per LED, connect Anode to +5V and place a resistor between the each LED Cathode and its related DTC113KZA  driver. 



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Note that the driver DTC113KZA has a 100 mA max load and a low current gain by most standards of only 33 minimum. (Max or typ not stated). It is OK for this application but would be poor for many output drive tasks.

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LED Current:

For LED on eg pin 1, LED current = current in R68.
I_R68 = V_r68/I_R68.
V_R68 = Vcc- VLED-Vdriver_on  

Say driver saturation is 0.2V.
Say I_LED_desired = 20 mA.
Look up LED datasheet - say (my guess)  V_LED = 1.9V. @ 20 mA.
Then V_R68 = 5 - 1.9 - 0.2 = 2.9V.
So R68 = V/I = 2.9V/20 mA = 145 Ohms. 

Change assumptions to suit.