Answer:
serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
     1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
     1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
     1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
     1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
     1 / Ceq = 1,147
     Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
    Dv = Q / Ceq
    Q = DV Ceq
    Q = 14 0.678 10⁻⁶
    Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
    ΔV = Q / C5
    ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
    ΔV = 1,725 V
d) The energy stores is
     U = ½ C V²
     U = ½ 0.678 10-6 14²
     U = 66.4 10⁻⁶ J
e) Parallel system
    Ceq = C1 + C2 + C3 + C4 + C5
    Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
    Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
    Q5 = C5 V
    Q5 = 5.5 10⁻⁶ 14
    Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
    U = ½ C V²
    U = ½ 22.7 10-6 14²
    U = 2.2 10⁻³ J