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I really do not get R11 between Q2's base and (near) Q3's emitter. ... How is Q2 biased anyway?
This is THE slickest way to bias two transistors. 95% of all 2 or 3 transistor phono preamps did it this way; also a ton of Pro gear from inputs to outputs.
Where can Q1 get any base bias current except through R11?
What is Q2's current here? (Assume Si transistors.)
We should know that if the circuit biases itself in a linear mode, Q1 base-emitter voltage is sure to be in the 0.5V-0.7V zone.
Let's pretend the resistor from Q2 emitter to Q1 base drops "no" voltage.
Then Q2 emitter must be at 0.5V-0.7V, That is the only way it will work. Starting from scratch, you can't be sure it will work; but we know the G1176 does work.
If Q2 emitter is at 0.5V-0.7V and has 1K, then Q2 current is obviously 0.5mA-0.7mA.
If Q1 has a large collector load resistor, its current is low and its Vbe is probably 0.5V. Therefore we refine Q2 current as 0.5mA.
If Q2 is loaded in 22K, the 22K at 0.5mA drops 11V. If the supply is +20V, the collector will sit at 20V-11V= 9V, and will be able to swing down near 1V and up near 19V (depending what comes after). This is a useful swing.
You have to understand this basic building-block first.
In the actual G1176, there's more stuff. Notably Q2 emitter is jacked up 0.51V by the DC path from Q3 emitter. And Q3 emitter resistor (the part Q1 base sees) is almost double. These two changes give nearly the same result as my example: 0.63mA.
We must check the "no voltage" assumption about the first transistor's base bias resistor. Two ways: look at collector current, know that base current is much smaller, and compare the resistors R12 and R11. Second base is at just a few volts, so most of the 25V appears on R12. R11 voltage drop is less, because it is only 560K say half, and mostly because the first transistor hFE is like 300. 23V/(2*300) is 0.04V drop in R11. This is small compared to the 0.5V Vbe and other bias factors. (Which is good, because real hFE may be 100 or 500; it's all small, so no great change.) _OR_ since we have observed voltages, compute the 0.63mA in 1.8K and compare to observed 1.02V at the other end of R11. This says more like 0.11V.... but the observation was taken with a 10Meg-20Meg meter, which will load the 0.56Meg R11 significantly. The student should compute the approximate meter loading for hisself.
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And the emitter of Q4 linked with Q2's one through R21 and R16?
What is the voltage gain through the three transistors? Depending on the actual parts used (hFE will matter) it may be 5,000 or 200,000. Different cold or hot or unit to unit. What gain do we want? Assuming 0.1V max at FET and up to 5V out, a total gain of 50 which MUST be uniform day-day unit-unit. The Q5-Q9 stage has gain of 3, so we need gain of 20 in the Q2-Q4 chain, fixed by good resistors not sloppy transistors. R21 and R13R14 are NFB to force the gain to 20. This is another basic connection you MUST absorb. Since worst-case Q2-Q4 have open loop gain of 5K, 250 times more than the NFB forces, the error will be under 1%.
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What about C2?
Ah, well, here's a trick. FET conductivity depends on voltage from gate to either end. In amplifiers one end (drain) has large voltage. But here both ends are at nearly the same voltage except one end is wiggling up and down with signal. So the average gate-channel voltage varies with signal. And conductivity varies with signal. And one side of the wave is shunted more than the other. The trick is to mix a little signal into the gate-channel voltage. R13 R14 R7 R9 form a bridge which I will not pretend to explain, but I assume works to correct FET gate-channel conductivity. (Specially since that's what R16 is supposed to fine-trim.)
