Electronic – Next level of resistor divider

No, the datasheet won't give you a minimum base current, because that's determined by the collector current you want. \$H_{FE}\$ is the parameter you need. For the BC557 (no suffix) that's minimum 125, which means that for a LED current of, say, 20mA you'll have

\$ I_B = \dfrac{I_C}{H_{FE}} = \dfrac{20mA}{125} = 160\mu A\$

If \$H_{FE}\$ would be higher (can be up to 800) the resistor in series with the LED will limit the current.

You can use a divider to set the base current, but this will cause unnecessary current parallel to the base-emitter junction. Just use a series resistor:

\$ R = \dfrac{\Delta V}{I} = \dfrac{5V - 0.7V}{160 \mu A} = 27k\Omega \$

That's the maximum resistance to get a guaranteed 20mA collector current. (I'm presuming your BC557 is connected to 5V, if it's 3.3V just redo the calculation.)

Note that the BC557 is a PNP transistor. If you want an NPN choose the BC547.

According to the product details, the output impedance is 100kΩ, so 100Ω would almost completely short the signal.
To get 1.2V from 3V, you need a ratio of 1.2 / 3 = 0.4

Assuming a steady 100kΩ output impedance, you will need a single 66.6kΩ resistor from the sensors output to ground to divide the range down as necessary:

3V * (66.6 / 166.6) = 1.2V

Looking at the xBee datasheet, the ADC input impedance is only 10kΩ, so an signal impedance of ~40kΩ (which is what you would have with the above) is not ideal, and you may lose accuracy.

Ideally, the best thing would probably be to use an simple non-inverting opamp placed in between this and the xBee ADC input to buffer the signal.