Some questions about a series pass transistor & op amp voltage regulator
\$R_1\$ and \$D_6\$ present a very basic zener voltage concept. You must become very, very familiar with it. It should be immediately obvious to you every time you see it. The basic idea is that the resistor, \$R_1\$ helps to "set the operating current" of the zener diode (and possibly supply a relatively much smaller current to something else attached at that node.) So if the zener diode is specified to operate at \$20\:\text{mA}\$, to just grab an example current out of the air, then you'd calculate the value of the resistor to provide about that current after subtracting the zener voltage from the supply voltage. The actual current you need is found on a datasheet. So that's the place you are supposed to read for it. (My example is only a "what if" example.)

That node shared by the zener diode (\$D_6\$) and the resistor (\$R_1\$) should be a relatively stable voltage that you can rely upon. This value is fed over to the (+) input of an opamp.

Separately, there is another very simple concept you must understand: the voltage divider which is created by \$R_3\$ and \$R_4\$. This divided voltage is then presented to the (-) input of the opamp.

You need to understand why the zener voltage is presented to the (+) input and the divided voltage is presented to the (-) input. It's pretty simple, really. If whatever is presented to the (+) input rises, then the output will rise. Oppositely, if whatever is presented to the (-) input rises, then the output will fall.

So, if the divider voltage falls, then the value presented to the (-) input will also fall and this will cause the output of the opamp to rise, in response. Since the emitter of the BJT follows the base voltage and since the base voltage equals the opamp output, this means that the opamp output will rise causing the emitter to also rise, therefore causing the voltage output to rise and therefore the divider voltage to rise in response. In other words, if the output tries to decline, the response of the system is to pull upward on the BJT emitter thus raising the output voltage to counter this change. Exactly what is desired.

Similarly, if the zener voltage itself were to rise (we supposedly "trust" the zener voltage), then this means the (+) input rises and therefore the opamp output rises, too. That pulls up on the emitter of the BJT causing the output to increase. Which is exactly what should happen. The output voltage follows the zener voltage.

So the inputs are correctly connected up.

The opamp just provides a lot of voltage gain. So any slight difference between the (-) and (+) inputs is greatly magnified at the output. This helps to keep the two inputs of the opamp very, very close in value to each other. Close enough, in fact, that you can consider them "virtually equal" to each other.

I'll leave the rest for you to work out. I think the above discussion, if you think closely about it, will lead you to a good answer for your third question. (As well as the other two.)