Answer
See the detailed answer below.
Work Step by Step
a) As we see below, at point a the gravitational force exerted on the satellite is perpendicular to the velocity vector direction.
So that the torque exerted by this gravitational force due to the planet's pull is zero.
$$\sum\tau=R_{orbit}\times F_G=R_{orbit}F_G\sin 0^\circ=\color{red}{\bf0}\;\rm N\cdot m$$
____________________________________________
b) It is obvious that the angular momentum of the satellite is conserved.
Thus,
$$L_a=L_b$$
$$I_a\omega_a= I_b\omega_b$$
Recalling that $\omega=v/R$ and that the moment of inertia of the satellite is given by $mR^2$
$$\color{red}{\bf\not} mR_av_a= \color{red}{\bf\not} mR_bv_b$$
Hence,
$$v_b=\dfrac{ R_av_a}{ R_b}$$
From the geometry of the given figure, we can see that $R_b=15,000+9,000=24,000\;\rm km$ and $R_a=15,000-9,000=6,000\;\rm km$.
So that,
$$v_b=\dfrac{ (6,000\times 10^3)(8000)}{ 24,000\times 10^3}=\color{red}{\bf 2000}\;\rm m/s$$
____________________________________________
c) By the same approach;
$$L_a=L_c$$
$$I_a\omega_a= I_c\omega_c$$
$$\color{red}{\bf\not} mR_av_a= \color{red}{\bf\not} mR_cv_c\sin\theta$$
where, from the geometry of the last figure below,
$\sin\theta=\dfrac{12,000}{R_c}$
$$ R_av_a= \color{red}{\bf\not} R_cv_c\dfrac{12,000}{\color{red}{\bf\not} R_c}$$
Hence,
$$v_c=\dfrac{ R_av_a}{ 12,000}$$
So that,
$$v_c=\dfrac{ (6,000\times 10^3)(8000)}{ 12,000 }=\color{red}{\bf 4000}\;\rm m/s$$
