Explanation

(a) Effective capacitance of $C_{2}\ and\ C_{3}$
$\ \ \ \ \ \ \ \ \ \ \frac{1}{C} = \frac{1}{2} + \frac{1}{2}$
$\therefore\ \ \ \ \ \ C = \ 1\mu\ F$
Now, $C_{1}and\ C$ are in parallel, therefore effective capacitance $C'$
$\ \ \ \ \ \ \ \ \ C^{'} = \ 1 + 1 = 2\mu\ F$
Now, $C^{'}and\ C_{4}$ are in series, therefore,
effective capacitance between points $A\ and\ B$
$\ \ \ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
$\Rightarrow \ \ \ \ C^{''} = \frac{4}{3}\mu F$