Answer
(a) $C_{bc} = 20 \,\text{pF}$
(b) $C_{ac} = 8.6 \,\text{pF} $
Work Step by Step
(a) Between $b$ and $c$, we have two capacitors are connected in parallel so its equivalent capacitance is given by
$$C_{bc} = 11 \,\text{pF} + 9\,\text{pF} = \boxed{20 \,\text{pF}}$$
(b) Between $a$ and $c$, we have two capacitance connected in series $C_{bc}$ and $15 \,\text{pF}$, so in parallel, thier eqivalent capaciatnce is given by
$$\dfrac{1}{C_{ac}} = \dfrac{1}{C_{bc} } + \dfrac{1}{15\,\text{pF} } = \dfrac{1}{20 \,\text{pF} } + \dfrac{1}{15\,\text{pF} } = 0.117 $$
Take the reciprocal
$$\boxed{C_{ac} = 8.6 \,\text{pF} }$$
