Answer
(a) Going from the earth to Mars, the rockets have to be fired in the direction opposite to the direction of motion to speed up the spacecraft at both points mentioned. Going from Mars to the earth, the rockets have to be fired in the direction of motion to slow the spacecraft down at both points mentioned.
(b) $259 \, \mathrm{days}$
(c) $44.1^{\circ}$
Work Step by Step
(a) First, the spacecraft has to speed up in order to escape the small circular orbit of the earth around the sun and enter the elliptical orbit. Then, upon getting into a position tangential to the orbit of mars, it would have to speed up again to exit the elliptical orbit and enter the larger circular orbit of mars. Thus, the rockets would have to be fired in the direction opposite of the motion of the spacecraft, at both points.
When going from Mars to the earth, the spacecraft first has to slow down in order to drop out of the large circular orbit of Mars and enter the elliptical orbit, then it has to slow down again upon getting tangent to earth's circular orbit in order to enter it. Thus the rockets have to be fired in the direction of motion at both times.
(The terminology used here is slightly confusing, but the key point is to understand when and where the spacecraft has to speed up/slow down.)
(b) The time it takes to get from the earth to Mars is half of the orbital period $T$ of the elliptical orbit with semi-major axis $a$, where $a$ can be found in the picture as the sum of Mars' and the earth's orbital radii divided by $2$: $a = (R_\mathrm{orb-M} + R_\mathrm{orb-E})/2$. Using the values given by Appendix F in the book, I have
$$\frac{T}{2} = \frac{2 \pi a^{3/2}}{2\sqrt{GM_\mathrm{S}}} = \frac{\pi \big((R_\mathrm{orb-M} + R_\mathrm{orb-E})/2 \big)^{3/2}}{\sqrt{GM_\mathrm{S}}} = $$
$$ \frac{\pi \big((2.28 \times 10^{11} \, \mathrm{m} + 1.50 \times 10^{11} \, \mathrm{m})/2\big)^{3/2}}{\sqrt{(6.67 \times 10^{-11} \, \mathrm{Nm^2/kg^2})(1.99 \times 10^{30} \, \mathrm{kg})}} = $$
$$ 2.241 \times 10^{7} \, \mathrm{s} = 259 \, \mathrm{days} \quad (259.322...) $$
(c) To find the angle between a sun-Mars line and a sun-earth line, I need to consider where Mars is at a time $T/2$ before it arrives in the right spot at the aphelion of the elliptical orbit of the spacecraft. Or rather, through how great an angle $\alpha$ it needs to travel to get to this position. Since I can assume that it moves with constant speed, the ratio of $T/2$ to the orbital period of Mars $T_\mathrm{M}$ is the same as the ratio of $\alpha$ in radians to a complete revolution:
$$\frac{T/2}{T_\mathrm{M}} = \frac{\alpha}{2\pi} \Rightarrow \alpha = \frac{2\pi (T/2)}{T_\mathrm{M}}$$
The angle $\beta$ between the earth and Mars at the time of launch is then
$$\beta = \pi - \alpha = \pi - \frac{2\pi(259.3 \, \mathrm{days})}{687.0 \, \mathrm{days}} = 0.770 \, \mathrm{rad} = 44.1^{\circ}$$
