Need help calculating resistor values for IR LED Driver
The LED resistor is calculated as
$$R_{LED}=\frac{V_{IN}-V_{F}}{I_F}$$

Which is, for a 200mA load:
$$\frac{5-1.2}{0.2} = 19\Omega$$

Pick the next highest common (E24) resistor, which would be 20Ω.

For a 100mA load the calculation would be:

$$\frac{5-1.2}{0.1} = 38\Omega$$

39Ω is the next highest E24 resistor, so that would do nicely.

You should also check the power handling requirements of the resistor.  Power can be calculated as
$$P=I^2R$$

Which for your 100mA load would be:
$$0.1^2×39 = 0.01×39 = 0.39W$$
So you should be using a half watt resistor, not the more common quarter watt resistors.  Alternatively, you can use multiple higher value quarter watt resistors in parallel to both achieve the small 39Ω resistance and the increased power handling.

The base resistor is there to limit the amount of current drawn through the IO pin.  The exact value in this situation is fairly unimportant.  Common values to use vary between 470Ω and 1KΩ - whatever you have in that region to hand really.  As the LED is using such low current you're not going to be needing to calculate the transistor's current gain or anything fancy like that.