With 0 on Q1's base, Q1 will saturate to probably within 100mV of the 5V rail i.e about 4.9V on the collector. If it were an NPN emitter follower and you applied 5V to the base, the emitter would be at 4.3V - maybe that's where you are having a minor bit of confusion?
With Q2's base sitting at 4.9V and its collector being at (say) 3.3V you'll have a heavily forward biased base-collector diode (un-natural operations) and may indeed damage Q2 due to excessive current flow.
Iif you had (say) a 10k resistor in series with Q2's base this would protect the device and you would see a voltage of nearly 3.3V on the emitter because Q2 would be saturated-on.
One, that switch does not directly control the motor. It's most likely a few mA at best, as it signals a microcontroller inside the cdrom to open/close the tray.
Two, what you are looking for is simple ohms law. Resistor = (Source Voltage - The Transistor Base/Emitter Drop) / Current Required in Amps. Since the hFE or current multiplication ability of the TIP120 is 1000, so roughly it will allow 1000 times the base current at the collector, any amount of current should be good at the base. Let's just use 5mA. The Base/Emitter drop is 1.25V minimum, as there are two transistor diodes.
Resistor = (5v - 1.25V) / 0.005A or 3.75V / 0.005A = 750Ω or close.
Update To further answer the question, you calculate the base resistor within the safe range of your source (Arduino, 40mA per pin, 200mA total at any given time). The unknown collector current in this case would be minimal for a button. For actual loads like a motor, you could simply max out the transistor by saturating it, giving the base transistor as much current as possible. In this case, you would have to have multiple arduino pins in parallel since the TIP120 base limit is 120mA and the Arduino is 40mA per pin. This is not ideal because you don't know the current at C-E or the amount of voltage it will drop.
The best answer is that you DONT. A proper design will find out how much current will be at the collector. Use a ammeter or multimeter in current mode to find out.
Best Answer
I think you need to jump (maybe not in at the deep end but) in a bit deeper and I would recommend studying the following graphs for a BC547 transistor. The data sheet is here: -
The top-left graph shows quite a few things: -
The top-right graph indicates what voltage can be expected to be seen on the base for driving a certain collector current with collector-emitter voltage held at 5V.
Bottom-left indicates how much current gain you can expect from the device.
Bottom-right is another representaion of what happens in the saturation region.
The trick here is asking the question how the collector current will vary with base current and once you have that you can decide what volt-drop will be across the resistor. From this you can check that the transistor is not entering saturation (because that puts you in the linear parts of the curves where collector current is more defined by collector voltage) and you'll need to re-appraise what the collector current is to recalculate the resistor volt-drop (iterations needed).
The bottom line is - try to fully understand the top-left graph - imagine a base current of (say) 200uA - ask yourself what collector current flows when the CE voltage is between 4V and 14V - the answer is about 50mA - hence the BJT is generally regarded as producing a constant current for a given base current but, the reality/detail is that collector current will vary between about 47mA (4V on collector) and 53mA (14 volt on collector). You might also start to notice that all the "constant-current" sections have an ever increasing slope and they all "point" to a position on the X-axis that is the same - it's called the Early voltage but a little too much to go into at the moment!!
Try to get a feel for this and recognize the onset of saturation and how it reshapes what I've just said - you can see, that for lower base drives, saturation is entered when collector voltage is less than 1V but, at higher base drives, saturation is entered at 2V or more - there is no exact definition of saturation that can be derived from the graphs so you just have to develop a gut-feeling for what it means.