Determine the equivalent capacitances between the specified points...

For circuit (a):
Let the input node be at potential $V$ and the output node be at $0$. Let the potentials at the top and bottom junctions be $x$ and $y$ respectively.
Applying Kirchhoff's Current Law (KCL) at the top junction:
$(x - V) \times 1 + (x - 0) \times 3 + (x - y) \times 4 = 0 \Rightarrow 8x - 4y = V$
Applying KCL at the bottom junction:
$(y - V) \times 3 + (y - 0) \times 1 + (y - x) \times 4 = 0 \Rightarrow -4x + 8y = 3V$
Solving these two equations, we get $x = \dfrac{5V}{12}$ and $y = \dfrac{7V}{12}$.
The total charge $Q$ supplied by the source is the sum of charges on the input capacitors:
$Q = (V - x) \times 1 + (V - y) \times 3 = \left(V - \dfrac{5V}{12}\right) \times 1 + \left(V - \dfrac{7V}{12}\right) \times 3 = \dfrac{7V}{12} + \dfrac{15V}{12} = \dfrac{22V}{12} = \dfrac{11V}{6}$
The equivalent capacitance is $C_{eq} = \dfrac{Q}{V} = \dfrac{11}{6} \mu\text{F}$.

For circuit (b):
Let the input be at $V$ and output at $0$. Let the potentials at the top, middle, and bottom junctions be $V_C$, $V_D$, and $V_E$ respectively.
Applying KCL at the three junctions:
Top: $(V_C - V) \times 1 + V_C \times 3 + (V_C - V_D) \times 4 = 0 \Rightarrow 8V_C - 4V_D = V$
Middle: $(V_D - V) \times 2 + V_D \times 2 + (V_D - V_C) \times 4 + (V_D - V_E) \times 4 = 0 \Rightarrow 12V_D - 4V_C - 4V_E = 2V$
Bottom: $(V_E - V) \times 3 + V_E \times 1 + (V_E - V_D) \times 4 = 0 \Rightarrow 8V_E - 4V_D = 3V$
Adding the top and bottom equations gives $8(V_C + V_E) - 8V_D = 4V \Rightarrow V_C + V_E = V_D + \dfrac{V}{2}$.
Substituting this into the middle equation gives $12V_D - 4\left(V_D + \dfrac{V}{2}\right) = 2V \Rightarrow 8V_D = 4V \Rightarrow V_D = \dfrac{V}{2}$.
Using $V_D = \dfrac{V}{2}$ in the top and bottom equations yields $V_C = \dfrac{3V}{8}$ and $V_E = \dfrac{5V}{8}$.
Total charge $Q = (V - V_C) \times 1 + (V - V_D) \times 2 + (V - V_E) \times 3 = \dfrac{5V}{8} + V + \dfrac{9V}{8} = \dfrac{22V}{8} = \dfrac{11V}{4}$.
The equivalent capacitance is $C_{eq} = \dfrac{Q}{V} = \dfrac{11}{4} \mu\text{F}$.

For circuit (c):
The upper part of the circuit forms a Wheatstone bridge. The ratio of capacitances in the top branch is $\dfrac{2}{4} = \dfrac{1}{2}$ and in the middle branch is $\dfrac{4}{8} = \dfrac{1}{2}$.
Since the bridge is balanced, no charge flows through the $5 \mu\text{F}$ capacitor, and it can be removed.
The equivalent capacitance of the bridge is $\dfrac{2 \times 4}{2 + 4} + \dfrac{4 \times 8}{4 + 8} = \dfrac{4}{3} + \dfrac{8}{3} = 4 \mu\text{F}$.
This is in parallel with the bottom $4 \mu\text{F}$ capacitor, so $C_{eq} = 4 + 4 = 8 \mu\text{F}$.

For circuit (d):
The circuit consists of four parallel branches. The ratio of capacitances in each branch is $\dfrac{2}{4} = \dfrac{1}{2}$, $\dfrac{4}{8} = \dfrac{1}{2}$, $\dfrac{4}{8} = \dfrac{1}{2}$, and $\dfrac{2}{4} = \dfrac{1}{2}$.
Because the ratios are identical, the midpoints of all four branches are at the same potential. Thus, the three $6 \mu\text{F}$ capacitors have no potential difference across them and can be removed.
The equivalent capacitance is the sum of the series combinations of each branch:
$C_{eq} = \dfrac{2 \times 4}{2 + 4} + \dfrac{4 \times 8}{4 + 8} + \dfrac{4 \times 8}{4 + 8} + \dfrac{2 \times 4}{2 + 4} = \dfrac{4}{3} + \dfrac{8}{3} + \dfrac{8}{3} + \dfrac{4}{3} = \dfrac{24}{3} = 8 \mu\text{F}$.

Answer: (a) $\dfrac{11}{6} \mu\text{F}$, (b) $\dfrac{11}{4} \mu\text{F}$, (c) $8 \mu\text{F}$, (d) $8 \mu\text{F}$