Mechanics question on coefficient of friction 

I am a bit stuck on this question, particularly part b. I have found that air resistance is equal to 689N from part a (though I am not sure that this is correct)
So far for part b, I have used suvat to find the deceleration of the car to be 6.05ms^2 and then used F=ma to find the force to be 13007.5 N but I don't know what this force is or if these steps are necessary. 
 A: For part a, when they say the $700$ N force acts in the direction of travel of the car, I am fairly sure they mean exactly in the same direction, that is, down the slope, not horizontally.
Also in part a, not only is the engine driving the car down the slope, there is also the effect of gravity.
For both those reasons, in order for the car to travel at a constant speed down the slope (zero net force), the air resistance has to be greater than what you have calculated so far.
For part b, you have computed the net force on the car during the time it is sliding to a stop. As labeled in your diagram, this is the sum of the air resistance and the frictional force. So if you compute the air resistance correctly the rest should just be applying the diagram and formulas you already have.
A: In the first situation, the forces involved are the following:

Applying the second Newton's law in the directions perpendicular and parallel to the motion:
$$
N - m\,g\,\cos\alpha = 0\,, \; \; \; F_d + m\,g\,\sin\alpha - F_a = m\,a
$$
and given that $a = 0$, we get:
$$
N = m\,g\,\cos\alpha = 20771.1\,N\,, \; \; \; F_a = F_d + m\,g\,\sin\alpha = 4362.5\,N\,.
$$
In the second situation, the forces involved are the following:

Applying Newton's second law parallel to motion:
$$
m\,g\,\sin\alpha - F_a - \mu\,N = m\,a
$$
i.e.
$$
m\,g\,\sin\alpha - \left(F_d + m\,g\,\sin\alpha\right) - \mu\left(m\,g\,\cos\alpha\right) = m\,a
$$
from which:
$$
a = - \frac{F_d + \mu\,m\,g\,\cos\alpha}{m}\,.
$$
On the other hand, the stopping time is equal to:
$$
0 = v_0 + a\,t 
\; \; \; \Leftrightarrow \; \; \; 
t = -\frac{v_0}{a}
$$
and given that the stopping distance is equal to $L$, we have:
$$
L = 0 + v_0\,t + \frac{1}{2}\,a\,t^2 = - \frac{v_0^2}{2\,a} 
\; \; \; \Leftrightarrow \; \; \;
a = - \frac{v_0^2}{2\,L}\,.
$$
Well, all that remains is to match the two expressions of acceleration:
$$
- \frac{F_d + \mu\,m\,g\,\cos\alpha}{m} = - \frac{v_0^2}{2\,L}
$$
from which it's possible to obtain the friction coefficient:
$$
\mu = \frac{m\,v_0^2 - 2\,F_d\,L}{2\,m\,g\,L\,\cos\alpha} = 0.592531\,.
$$
