A + 4KOH + O2 → 2B (Green) + 2H2O
3B + 4HCl → 2C (Purple) + MnO2 + 2H2O
2B + H2O + KI → 2A + 2KOH + D
In the above sequence of reactions, A and D respectively, are:

The given reactions are:

A + 4KOH + O2 → 2B (Green) + 2H2O
3B + 4HCl → 2C (Purple) + MnO2 + 2H2O
2B + H2O + KI → 2A + 2KOH + D

Let's analyze the reactions:

Reaction 1: A + 4KOH + O2 → 2B (Green) + 2H2O
This reaction suggests that A is oxidized by O2 in alkaline medium to form a green compound B.

Reaction 2: 3B + 4HCl → 2C (Purple) + MnO2 + 2H2O
This reaction indicates that B is further reacted with HCl to produce a purple compound C, MnO2, and water. This suggests that B is likely a manganese compound.

Reaction 3: 2B + H2O + KI → 2A + 2KOH + D
This reaction suggests that B is reduced by KI in the presence of water to reform A and produce compound D.

Considering the oxidation states of manganese, the reactions can be identified as follows:

Reaction 1: 2KI + 4KOH + O2 → 2K2MnO4 (Green) + 2H2O (A = KI)
Reaction 2: 3K2MnO4 + 4HCl → 2KMnO4 (Purple) + MnO2 + 4KCl + 2H2O (B = K2MnO4, C = KMnO4)
Reaction 3: 2K2MnO4 + H2O + 2KI → 4KOH + 2KI + 2MnO2 (B = K2MnO4, D = MnO2)

Therefore, A is KI and D is MnO2.

Final Answer: The final answer is $oxed{KI and MnO2}$