Voltage Divider to Power LED
This is a homework problem so I will only point towards an approach...
Homework problems are usually constructed around some specific teaching point, in this case I think they're trying to teach &quot;A rule of thumb is that the current through the voltage divider is 10 times as large as that which passes over the consumer&quot;. That is in fact a good rule of thumb. So that sets the bias current needed through R1 and R2, based on how much current needs to drive the LED. (As a side note: be sure to think about how much power will be lost in R1 and R2.)
This is a very bad circuit for driving a motor, because it lacks the freewheeling diode to protect against the voltage surge which happens when the motor turns off and the rotor is still spinning. If you haven't learned about inductors yet, be warned, inductors respond to a change in current, with an opposing change in their terminal voltage. So switching off an inductor causes a sharp change in voltage. How much depends on how rapidly the current is changed. But this is beyond the scope of this homework problem. Just keep in mind this isn't a real motor driving circuit...
This is also a very bad circuit for driving an LED, because the voltage divider R1/R2 already performs the role of limiting the LED drive current, so R3 is not really necessary. Most likely R3 was put there because of a &quot;design rule&quot; that an LED always requires a series resistor (not strictly true BTW, just a good rule of thumb). R3 just adds a needless degree of freedom which isn't required in this case, and just gets in the way of solving the problem
If I were choosing the series resistor for an LED, I would need to know the LEDs operating forward voltage \$V_{f}\$ and forward current \$I_{f}\$, and since I know both the voltage at each end of the resistor and the current through the resistor, I would just apply Ohm's law. But, as noted in a comment by @Justme, this badly designed circuit has an error. As drawn, they've specified Udiod=3.5V, which is apparently also meant to be the forward voltage of the LED. So there is no voltage drop left for R3. Either that, or you're expected to arbitrarily choose the voltage at both ends of R3, and the circuit would be undecidable. As a practical matter, 3.5V is a credible value for an LED forward voltage (red LEDs are usually about 1.5V@10mA, other colors often require higher forward voltage); so my best guess is that R3 is a mistake and should be replaced by a direct &quot;short&quot; connection to LED1.
I don't know whether you've learned Thevenin/Norton equivalent sub-circuits yet, but just FYI the 12V voltage source with resistive divider R1/R2, is equivalent to a voltage source of 12V x R2 / (R1 + R2) with a series resistor of (R1 parallel with R2). So the R1/R2 voltage divider's equivalent parallel resistance is what satisfies the design rule of connecting a resistor in series with an LED to limit its current.
R3 just adds a needless degree of freedom which isn't required in this case, and just gets in the way of solving the problem. So my approach would be to set R3=0 and then determine R1/R2 based on the required voltage and current to the LED. However @Neil_UK in the comments suggested another approach which discards R1/R2 by setting R1=0 and R2=not installed, and then determines a value for R3 using the approach I described earlier, 12V on one end of R3, 3.5V on the other end of R3, and applying Ohm's law. Both are valid approaches to this badly constructed homework problem.
A much better textbook for learning analog electronics is The Art of Electronics (Horowiz and Hill, Cambridge University press, ISBN 0-521-37095-7).