Your answer for part a is correct. Using Newton's 3rd Law, the force on the rocket by the exhaust leaving the rocket is the same magnitude (opposite direction) as the downward force applied to the exhaust.
In part b the net force is the upward force from the exhaust thrust minus the downward force of gravity:
[tex]F_{net}=F_{thrust}-mg=15,000N-1500kg \times 1.6 \frac{m}{s^2}=12,600N [/tex]
Then using the Second Law, we get for part c:
[tex]F=ma \\ a= \frac{F}{m} = \frac{12600N}{1500kg} =8.4 \frac{m}{s^2} [/tex]
The force and acceleration are in the upward direction
