A V/Q mismatch can produce Hb with less than the normal amount of O2 molecules attached, but this has nothing to do with cooperative kinetics. I think you're mixing up multiple concepts. Shunted blood, eg, won't reach 100% sat because its not exposed to high O2 tension; not because of cooperative kinetics.
Hb has four possible states, since it can bind 4 O2 molecules. Hb passing through the lung will hit 100% sat, but a small amount remains at venous sat (75%), because theres a small shunt across the lungs even under normal circumstances. This Hb remains at lower sat because it isn't exposed to the gas exchanging areas of the lung. It has nothing to do with cooperative kinetics, which is the source of your confusion, and mine, since I assumed it was central to this discussion. This small shunt hitting the arterial circulation will then decrease the net saturation of the blood to be a bit under 100%.
If you were to pull a Hb molecule at random from arterial blood that has yet to supply tissue, the overwhelming odds are that it will have 4 O2 molecules attached. Passing through tissues, Hb molecules with 4 O2 attached will give up one O2 and hit 75%. The small fraction that was shunted across the lungs and was already at 75% before reaching tissues, won't give up any O2, since tissue tension isn't low enough for a second O2 to be lost. And if it were, all Hb molecules would be equally prone to hitting 50%.
The blood entering the lungs will be at 75%, and the cycle repeats.