Answer
See the figures below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To reach a net capacitcance of $4\;\rm\mu F$, we should connect the three $12\;\rm\mu F$-capacitors in series.
$$C_{eq}=\left[ \dfrac{1}{12}+ \dfrac{1}{12}+ \dfrac{1}{12}\right]^{-1}=\bf 4\;\rm \mu F$$
See the first circuit below.
$$\color{blue}{\bf [b]}$$
To reach a net capacitance of $8\;\rm\mu F$, we should connect two of them in parallel and the last one should be in series to this parallel combination.
$$C_{eq}=\left[ \dfrac{1}{24}+ \dfrac{1}{12} \right]^{-1} =\bf 8\;\rm \mu F$$
See the second circuit below.
$$\color{blue}{\bf [c]}$$
To reach a net capacitance of $18\;\rm\mu F$, we should connect two of them in series and the last one should be in parallel to this series combination.
$$C_{eq}=\left[ \dfrac{1}{12}+ \dfrac{1}{12} \right]^{-1}+12=\bf 18\;\rm \mu F$$
See the third circuit below.
$$\color{blue}{\bf [d]}$$
To reach a net capacitance of $36\;\rm\mu F$, we should connect the three $12\;\rm\mu F$-capacitors in parallel.
$$C_{eq}=12+12+12=\bf 36\;\rm \mu F$$
See the fourth circuit below.
