Answer
The answer is shown below.
Work Step by Step
(a) $S_1$ contains only charge $q_1$, so the electric flux through this surface is
$$\Phi_{S_{1}}=q_{1} / \epsilon_{0}=\left(4 \times 10^{-9} \mathrm{C}\right) / \epsilon_{0}=452 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$$
where $\epsilon_o = 8.854 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$
(b) $S_2$ contains only charge $q_2$, so the electric flux through this surface is
$$\Phi_{S_{2}}=q_{2} / \epsilon_{0}=\left(-7.8 \times 10^{-9} \mathrm{C}\right) / \epsilon_{0}=-881 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$$
(c) $S_3$ contains the charge $q_1$ and $q_2$, so the electric flux through this surface is
$$\Phi_{S_{3}}=\left(q_{1}+q_{2}\right) / \epsilon_{0}=\left((4-7.8) \times 10^{-9} \mathrm{C}\right) / \epsilon_{0}=-429 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$$
(d) $S_4$ contains the charge $q_1$ and $q_3$, so the electric flux through this surface is
$$\Phi_{S_{4}}=\left(q_{1}+q_{3}\right) / \epsilon_{0}=\left((4+2.4) \times 10^{-9} \mathrm{C}\right) / \epsilon_{0}=723 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$$
(e) $S_5$ contains the charges $q_1, q_2$ and $q_3$, so the electric flux through this surface is
$$\Phi_{S_{5}}=\left(q_{1}+q_{2}+q_{3}\right) / \epsilon_{0}=\left((4-7.8+2.4 ) \times 10^{-9} \mathrm{C}\right) / \epsilon_{0}=-158 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$$
(f) No, it doesn't depend because the Gauss's law is for the enclosed charge.
