Hi idtkid
5 IR-LEDs of Uf=1.28V connected in series will require a total voltage of 6.4V.
Using the 7.5V power supply there will be 1.1V left to "kill". Applying Ohms law the current limiting resistor must be (US-UF)/IF, (Supply voltage-LED forward voltage)/LED forward current))
(7.5V-6.4V)/0.03=36.333Ohm. Next higher standard resistor value is 39Ohm. Using that value the total LED current will decrease to 28.205mA, which is acceptable.
For the IR-camera use the same power supply and make a voltage regulator circuit using an LM317T (adjustable voltage regulator, 1A). The data sheet contains a suggestion for an adjustable power supply. Instead of using a potentiometer you might use a trimpot to obtain a fixed value output voltage of 3V.
Boncuk