here in this question's solution, given that upper BJT is on while lower one is in cutoff region for Vbb = 2.7 V and viceversa for Vbb = -2.7 V and both will be in cutoff for 0 V. While in other problem Q1 is on and Q2 is in cutoff.
So my question is how to decide which one will be conducting and which one will be off and why? Is there any general procedure?
Look at this circuit carefully and realize each transistor is being used in emitter follower mode. However, as you say, at most one of them is on at a time.
To decide whether a transistor is on or off, look at the voltage across its B-E junction.
Hint: Remove any transistor that is off from the circuit, then analyze.
Your two questions are not complicated. I suspect you are imagining they are worse than they really are. Let's apply simple logic with the more recently added schematic:
simulate this circuit – Schematic created using CircuitLab
First thing to do is to decide what the two transistors are doing. Suppose that \$Q_2\$'s \$V_{BE_2}\approx 700\:\textrm{mV}\$ and that it is therefore actively on. Then it follows that \$V_X\approx -2.7\:\textrm{V}\$. But if that were the case, then \$Q_1\$'s \$V_{BE_1}\approx 5.7\:\textrm{V}\$! Given that the collector current goes up by a factor of about 10 for each \$60\:\textrm{mV}\$, this would imply a HUGE collector current and probably a huge base current in \$Q_1\$. Not to mention that \$R_1\$ couldn't allow it, anyway.
Given that the two emitters share the same node, it's much more likely now that these details are exposed that \$Q_2\$ is actually off and \$Q_1\$ is the only transistor in an active mode. Once you realize this, you then know that \$V_X\approx 2.3\:\textrm{V}\$, that \$Q_2\$ is off, \$I_O=\frac{+2.3\:\textrm{V}- \left(-6\:\textrm{V}\right)}{2\:\textrm{k}\Omega}=4.15\:\textrm{mA}\$, \$I_{C_1}\approx 4.15\:\textrm{mA}\$ (slightly less than the emitter current because of a tiny base current), and that \$I_{C_2}\approx 0\:\textrm{mA}\$.
Another way to see that fact is to simply imagine the two transistors as both being completely off to start, and to then gradually activate \$R_1\$ by slowly pulling the \$-6\:\textrm{V}\$ rail downward, starting at the highest base voltage of \$+3\:\textrm{V}\$ and ramping it downward towards \$-6\:\textrm{V}\$ just a little bit at a time. Note that as you do this mental step, \$Q_1\$ will definitely be the first BJT to turn on. And once it turns on at around \$600\:\textrm{mV}\$ or so, when the bottom supply rail has only just reached about \$+2.4\:\textrm{V}\$, the emitter current will start rising upward by factors of 10 for each additional \$60\:\textrm{mV}\$ lower. It doesn't take much to then realize that the emitter of \$Q_1\$ will never allow the base-emitter junction of \$Q_2\$ to become forward biased. It just can't happen. So \$Q_2\$ is off. End of story.
Now you should be able to apply similar logic in your original case.
In cut off mode both junction of bjt (emitter-base and collector-base) will be in reverse biased condition. It can happen only if base is biased negatively (in NPN transistor) and biased positively in (PNP transistor).
