OK, this is kinda simpler than it looks
H801/H802 are in a Darlington arrangement where they act pretty much like a single high gain transistor, Assume that the collectors are the collector of the transistor, the emitter of H801 is the emitter, and the base of H802 is the base.
Power comes in through the fuse F601.
The transistor H801/2 is turned ON via R803 and R801
If just these components were in the circuit, the output (on the emitter of H801/2) would be essentially the full supply voltage (less losses in the transistor)
The purpose of H803 and the components generally to the right of it are to rob base current from H801/2 in order to partially turn H801/2 off.
Recall that an NPN transistor begins to turn on when current floes through the BE junction. This requires around 0.6V to overcome the the voltage drop through a diode. So, when the base of H803 gets more than 0.6V more positive than its emitter, (which is held at approx 7.4 volts by the zener) current will be robbed from the base of H801/2 and redirected to ground through H803.
The source of the voltage for the base of H803 is a voltage divider formed by Rs 805/806/807. If you assume the 5k pot (R806) is centred, the 2 halves of the voltage divider measure 24.5k and 7.85k. This means the voltage on the base of H803 is approximately 1/4 of the output voltage. The output voltage is thus kept to approx the (zener voltage + 0.6) * 4, or 32 volts. This matches well with the expected 35 volts (and also means R806 will not be centred)
All that's left are a couple of capacitors. In general they provide smoothing of the various voltage sources which (in the case of the first 2) will improve ripple rejection, and in the case of C807 will improve stability (I think)
Yes the voltage on the base of H801/2 will be approx 36.3 V (2 diode drops higher than the emitter)
The voltage drop across R801 and R803 should be the same. If they are different (the drop across R801 being higher) then I would suspect C805(?). It may be leaky. Try removing it and seeing if the voltages rise to their expected values. If they do, replace it with a new one.
If C807 is leaky, the voltage will be regulated through the full range of R806, but at a lower voltage. If this is the case, removing C807 should restore the output to the correct voltage, and replacing it with a new one will fix things.
A leaky C806 may cause the voltage to be adjustable by R806, but having it fall under load or simply not be adjustable beyond some position of R806. Again, removing it will restore the voltages, and if this occurs, replacing it will fix the problem.
None of those capacitors look like they have a particularly hard life. If they are close to heatsinks then they may have been damaged over time by heat. Or they just might be old and leaky.
It may simply be easier to replace the three capacitors and see if the problem is resolved, or you may wish to individually check each one. If any of them are at fault, I would be inclined to replace them all.