Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the plane bisects the dipole which means that the plane is perpendicular to the dipole.
The author told us that the distance between the charge $Q$ and the dipole is much greater than the length of the dipole itself ($r\gt\gt s$).
So, from all of that, the electric field exerted by the dipole on the position of the charge $Q$ is given by
$$E_{dipole}= \dfrac{1}{4\pi \epsilon_0}\dfrac{p}{r^3} $$
where $p=qs$
$$E_{dipole}= \dfrac{1}{4\pi \epsilon_0}\dfrac{qs}{r^3} \tag 1$$
And hence, the force exerted by the dipole on the charge is given by
$$F=QE_{dipole}= \dfrac{Q}{4\pi \epsilon_0}\dfrac{qs}{r^3}$$
According to Newton's third law, the force exerted by the charge $Q$ on the dipole is equal in magnitude to the force exerted by the dipole on the charge $Q$.
$$\boxed{F_{\rm on\;dipole}= \dfrac{1}{4\pi \epsilon_0}\dfrac{Qqs}{r^3}}$$
And its direction is in the same direction of the dipole momentum $\vec p$.
$$\color{blue}{\bf [b]}$$
We know that the torque is given by
$$\tau=pE\sin\theta$$
We know that the electric field exerted on the dipole by the charge which is on a plane that is perpendicular to the dipole is also perpendicular the to the dipole. Thus $\theta=90^\circ\rightarrow\sin\theta=1$
$$\tau=pE =qsE_{\rm charge}$$
where $E_{\rm charge}$ is the electric field exerted by the charge on the middle of the dipole which is given by $kQ/r^2$.
Thus,
$$\tau_{\rm on\;dipole} =qs \left[\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{r^2}\right]$$
$$\boxed{\tau_{\rm on\;dipole} =\dfrac{1}{4\pi \epsilon_0}\dfrac{Qqs}{r^2}}$$
