That gain of 100 to 300 is for small signals only. If you want to drive a transistor to saturation, you'll need a lot more current than indicated from that. Gain goes down as the transistor reaches saturation, and you want the transistor to be fully On.
I usually use about a 10:1 ratio to fully saturate a 2N2222. So for 300mA, I'd push 30mA into the base.
This PDF specifies 15mA IBE for 150mA ICE with VCE(sat) of 0.3 to 0.4V, and VBE(sat) of 1.2 to 1.3V, depending on if it is a 2N2222 or 2N2222A.
Page 2:
http://www.aoc.nrao.edu/~pharden/hobby/HG_DS1.pdf
So 5V - 1.3V = 3.7V, 3.7V/30mA = 123 ohms so use R1 = 120 ohms
As for the current limiting resistor on the LEDs, I think you'll need to split the blue/green LEDs up into two parallel strings. That battery is going to vary in voltage quite a bit. If you don't leave enough headroom, the current in the LEDs will vary widely as the battery goes from fully charged to discharged. The red LEDs should be fine.
So for blue/green at 3.3V per LED, (13.8 - (2 x 3.3 + 0.4))/0.3 = 22.7 ohms or 22 ohms in two sets of 2 LEDs with one 22 ohm resistor on each string. That's about 6.8V of headroom.
For Red at 2.1V per LED, four in series is fine. (13.8 - (4 x 2.1 + 0.4))/0.3 = 16.7 ohms or 16 ohms. That is only 5V of headroom.
As the voltage drops from 13.8V to 12V, current in the blue/green LEDs will drop to about 73.5% or about 220mA, current in the red LEDs will drop to about 64% or about 190mA.