1) What will be the output voltage?
It will be 9 minus 0.7V = 8.3 V. Since it's an emitter follower configuration, the emitter side voltage will be always lower than the base by a magnitude equal to the forward drop of the base. The forward drop of a BJT is around 0.7V, and the base voltage is clamped at 9V, so the effective emitter voltage will be 8.3V.
2) What will happen to the LED and the transistor, since both are without a series resistor?
Since the emitter is the common exit terminals for both the LED current and the base current, the LED and the transistor both will be safeguarded by the 1K resistor. The LED will light up normally, and will switch OFF only if the base is disconnected from the positive line.
3) In this circuit the diode seems to be shorting the base of the transistor to ground, will the LED illuminate?
The LED will illuminate optimally because the diode has a forward voltage drop rating of 0.6 V, which is also the minimum required base switching voltage of the transistor. Therefore, the transistor base will not be completely at the ground potential, rather at 0.6 V above ground, which will be enough to switch it ON and the illuminate the LED fully.
4) This diagram is similar to the previous diagram, except the inclusion of an emitter diode. How will the LED respond now?
The transistor will remain shut off! Because the 0.6 V drop generated by the diode at the base of the transistor will be blocked or cancelled by the 0.6 V drop generated by the at the emitter diode of the transistor. In other words, the 0.6 V created at the the base of the transistor will not be able to connect with the ground through the emitter due to the presence of the another diode. The base will require over 0.6 + 0.6V to show any possible conductiviity.
5) One of the users notices that the base emitter voltage of a transistor circuit continues to be at 0.6 V, even though the input supply is varied from 3 V to 12 V. Is the transistor faulty?
The transistor is alright, and working normally. The 0.6 V is due to the base/emitter characteristic of BJTs which allows them to have a forward voltage drop of 0.6 V across base emitter. The 1 K base resistor drops the supply current sufficiently at the base which pulls the supply voltage down to the 0.6 V level.