When the collector of an NPN transistor is negative with respect to the emitter, the C-B junction is forward biased, so you're essentially applying a reverse voltage to the B-E junction. According to the datasheet, the maximum that this transistor can withstand is 5V (VEBO). For any voltage greater than this, you'll need to limit the current externally in order avoid damaging the transistor.
Yes, you can connect a diode across the transistor to bypass reverse current around it, limiting the reverse voltage that the transistor sees to the VF of the diode.
I've worked in electronics all my life (and am now retired). I perhaps did do calculations based in Kirchoff's laws, current division, etc in school, but now just consider Kirchoff's laws to be scientific wording of what should be simple, common-sense observations.
When I look at this problem, I see the 0.1 mA through the 60K resistor. I see that there is a resistor of twice that value (120K), in parallel, so it will have half the current of the 60K, so the total current in the circuit will be 0.15 mA.
The voltage across the 60K resistor is 60K x 0.1 mA, or 6 volts.
The voltage across the 20K resistor is 20K x 0.15 mA, or 3 volts, so Vs is 6 + 3 = 9 volts.