RGB LED powered with USB Questions
$$LED\;current = \frac{input\;voltage - LED\;forward\;voltage} {LED\;resistor}$$

$$I_{LED} = \frac{V_{USB} - V_{f_{LED}}} {R_{LED}}$$

Let us assume a current of 20mA for the LED.

$$0.02A = \frac{5V - 3.0V} {R_{LED}}$$
$$R_{LED} = 100\Omega$$

$$0.02A = \frac{5V - 3.6V} {R_{LED}}$$
$$R_{LED} = 70\Omega$$

I would use a 100 &Omega; LED.

At 20mA per LED, you could have \$ \dfrac{0.5A}{0.02A} = 25\$ LEDs in parallel. 

The kit is supplied with 200 &Omega; resistors. Using them you could have 50 LEDs in parallel.