The calculations below are performed assuming a vacuum, as the dielectric constant for kerosene is not provided. Let $q_1$ be at $x=0$, $q_2$ at $x=1.0$ m, and $q_3$ at $x=0.5$ m. The distance from $q_1$ and $q_2$ to $q_3$ is $r = d/2 = 0.5$ m. All charges are positive, so all forces are repulsive.

Force on $q_3$: The force from $q_2$ ($F_{23}$) is stronger than from $q_1$ ($F_{13}$). The net force is directed towards $q_1$.

$$F_{net,3} = F_{13} - F_{23} = k\frac{q_1 q_3}{r^2} - k\frac{q_2 q_3}{r^2} = \frac{kq_3}{r^2}(q_1 - q_2)$$ $$F_{net,3} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{(0.5)^2}(1.0 \times 10^{-8} - 2.0 \times 10^{-8}) = -1.08 \times 10^{-6} N$$

Force on $q_1$: Both $q_2$ and $q_3$ repel $q_1$ to the left.

$$F_{net,1} = -F_{21} - F_{31} = -k\frac{q_2 q_1}{d^2} - k\frac{q_3 q_1}{r^2}$$ $$F_{net,1} = -k q_1 \left( \frac{q_2}{d^2} + \frac{q_3}{r^2} \right) = -(9.0 \times 10^9)(1.0 \times 10^{-8})\left( \frac{2.0 \times 10^{-8}}{1^2} + \frac{3.0 \times 10^{-9}}{0.5^2} \right) = -2.88 \times 10^{-6} N$$

Force on $q_2$: Both $q_1$ and $q_3$ repel $q_2$ to the right.

$$F_{net,2} = F_{12} + F_{32} = k\frac{q_1 q_2}{d^2} + k\frac{q_3 q_2}{r^2}$$ $$F_{net,2} = k q_2 \left( \frac{q_1}{d^2} + \frac{q_3}{r^2} \right) = (9.0 \times 10^9)(2.0 \times 10^{-8})\left( \frac{1.0 \times 10^{-8}}{1^2} + \frac{3.0 \times 10^{-9}}{0.5^2} \right) = 3.96 \times 10^{-6} N$$

For [Q2], yes, the effect of kerosene (a dielectric medium) should be considered. It reduces the electrostatic force by a factor equal to its dielectric constant $\epsilon_r$. The forces calculated above would be smaller in kerosene.