Explanation

(a) The circuit diagram may be redrawn as shown here.
Obviously, $I_{CAD} = I_{CBD} = \frac{2}{12}A$
$\therefore\ V_{C} - V_{A} = \frac{2}{15}A \times 5\Omega = \frac{2}{3}V$
and $V_{c} - V_{B} = \frac{2}{15}A \times 10\Omega = \frac{4}{3}V$
$\therefore V_{A} - V_{B} = \left( V_{C} - V_{B} \right) - \left( V_{c} - V_{A} \right) = \frac{4}{3}V - \frac{2}{3}V = \frac{2}{3}V$