Infrared spotlight project, need x50 resistors. How many ohms what type to use?
This is an application of Ohm's law.

Assuming each set of three LED's are connected in series with a current limiting resistor:

Using a \$V_f\$ of 1.5 means the total \$V_f\$ for each diode group is 4.5 V.

The resistor value is \$R = \frac{E}{I}\$, or \$R = \frac{E}{0.060}\$


at 6V: \$R = \frac{(6 - 4.5)}{ 0.060} = \$ 25 ohm (90 mW, use 1/4 W resistor)
at 12V: \$R = \frac{(12 - 4.5)}{0.060} = \$ 125 ohm (450 mW, use 1/2 W resistor)


Be sure to use a resistor rated for the power dissipation (\$P = IE\$). Assuming these 50 groups are in parallel, that works out to be a total of 3 amperes. You'll lose less power by having smaller resistor from a lower supply voltage, so of these examples, 6 volts is preferable to 12 volts. To be more efficient you may want to research LED driver circuits.

Make sure your power supply/batteries can supply at least 3A!

Edit:

Per your edit, using 14.8 V would require larger current limiting resistors and waste more power:

14.8V: \$R = \frac{(14.8 - 4.5)}{0.060} = \$ 172 ohm (618mW, use 1W resistor)

At this voltage, the power dissipation of the resistor is now 618 mW and you would therefore have to use 1W power resistors.

Edit 2:

Some of the math and resistor values were off. It was late at night apparently, and I was trying to get the TeX right. I've adjusted the math and values to be accurate.