How to derive the node voltage for this comparator with hysteresis?
In the absence of contribtutions from R21, 

V(pin 3) = 5 x (56.2k + 10k)/(56.2k + 10k + 100k)

V(pin 6) = 5 x (56.2k)/(56.2k + 10k + 100k)

Furthermore, the hysteresis at pin 3 will be about 10 times greater than you think.

Including the effects of R21, and ignoring the effects of loading via the GREEN LED, 

for high input voltage, V(pin 3) = 5 x (56.2k + 10k)/(56.2k + 10k + 50k)

for low voltage, V(pin 3) = 5 x [(56.2k + 10k) x 100k) / [(56.2k + 10k + 100k)] /([(56.2k + 10k) x 100k) / [(56.2k + 10k + 100k)] + 100k)

Then, in both cases, V(pin 6) = V(pin 3) x 56.2k / (56.2k + 10k)

A number of suggestions -

1) Get rid of C14. It will likely cause stability problems for U26. If you must do some filtering, put it in parallel with R29.

2) Get rid of C21. Its low-pass filtering will interfere with the snap action that you want from the R21 positive feedback.

3) ETA: Your logic only works if your LEDs are connected anode to +5. If you try to drive to ground the logic reverses and you'll have 2 on at any time.

4) Divide R13, R14 and R16 by 100. Then insert a 2k resistor between the junction of R13/R14 and pin 3. Even better, don't cut the resistor values, and don't get rid of C21, but buffer the R13/R14 voltage with another op amp voltage follower just like you're doing with U26, and feed that to pin 3 with a 2k resistor.