A

Step 1: The potential difference across a capacitor is given by the relationship $V = \frac{Q}{C}$, where $Q$ is the charge and $C$ is the capacitance.
Step 2: The capacitance of a capacitor is given by $C = \frac{\varepsilon A}{d}$, where $\varepsilon$ is the permittivity, $A$ is the area, and $d$ is the separation between the plates. When the separation $d$ is doubled, the new capacitance becomes $C' = \frac{\varepsilon A}{2d} = \frac{C}{2}$.
Step 3: If the charge remains constant, the new potential difference across C1 will be $V' = \frac{Q}{C'} = \frac{Q}{C/2} = 2\frac{Q}{C} = 2V$. Thus, the potential difference increases.
Step 4: For capacitor C2, since the potential difference across parallel capacitors is the same, it will also experience a similar increase in potential drop, depending upon its own capacitance.
Step 5: Finally, the energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$. As capacitance decreases, the energy stored will decrease in C2 but increase in C1 after the distance is doubled.
Thus, the responses can be matched as:
(i) increases [A]
(ii) decreases [B]
(iii) increases by a factor of 2 [C]
(iv) decreases by a factor of 2 [D].
Therefore, the correct option is A.