The equation you're missing is:
f_o = f_s \cdot \frac{v \pm v_o}{v \mp v_s}
Where f_o is the observed frequency
f_s is the original frequency
v is the speed of sound
v_o is the speed of the observer
v_s is the speed of the object producing the sound.
The hardest part is getting the signs right with two moving objects. The best way of doing this is consider what would happen if each one of the objects was stationary in separate scenarios. The example my text has for this is a submarine traveling at 5 m/s detecting a signal from a submarine behind it traveling at an unknown speed. It's given in the problem that the frequency increases, so let's use Doppler logic to get the signs right.
First, let's assume the situation in which the observing sub is stationary. Given the difference in speeds, this means that the sub producing the sound is closing in. The Doppler effect says that the frequency would increase in such a situation. The only way for \frac{v}{v \mp v_s} to increase the frequency is if the denominator is less than the numerator, so we use a minus on the bottom.
For the top, if the sub producing the sound was stationary, the two subs would be drifting apart, meaning that the frequency would decrease. In order for the fraction to decrease the frequency, the numerator must be smaller than the denominator, so we use a minus on the top as well.
So, let's look at the problems:
a. you can use the above formula with the speed of the firetruck as v_s and the speed of the car as v_o to determine the answer.
b. Since the chirps reflect back off the wall, the bat is flying into the reflection. Since the sound and the receiver are coming together, the frequency will increase. A use of the above formula with an arbitrary speed of the bat will show this. However, the bat is also in the same location that the sound is produced so it is a bit more complicated since it is hearing both the reflection and the produced sound. You could say that, although the produced sound did not change frequency, the frequency of the reflection did increase.
c. If both forks are in the same room, the two waves will travel at the same speed because frequency and wavelength are proportional to each other by a constant speed of sound (v=\lambda \cdot f).
Hope that helped.
Steven
Edit: upon rereading your original post, it seems that these questions are over the general concept of the Doppler effect, rather than the mathematics behind it. In that case, let me explain that in the context of the problems:
a. Since the frequency of sound increases when objects come together and decreases when they diverge, whichever of the two objects the firetruck is approaching the fastest will hear the highest frequency. Using reference frames, the firetruck is approaching the driver of the van at 40 m/s and the driver of the car at 10 m/s, so the van will hear a higher frequency.
You were correct for b and c. Sorry for making the problems more complicated than they were.
