Poisson probability question For part a I think its just the poisson probability of 40 with rate = 48. I'm stuck on b to d.
Cars pass checkpoint A in accordance with a Poisson process at an average rate of 24
cars per hour. All cars passing checkpoint A also pass checkpoint B which is located 10
miles down the road. The speed (MPH) of any individual car between A & B is constant
and uniformly distributed on (40, 60). The speeds of all cars are mutually independent.
Find:
a) P(40 cars pass A between noon & 2PM)
b) P(between 3 & 6PM, 20 cars pass A at a speed exceeding 55 MPH)
c) P(10 cars pass B between noon & 1 PM)
d) the probability that 5 cars pass B between 1 PM & 1:10PM, given that
exactly 10 cars pass A between 12:30PM & 1 PM.
 A: [Much time has passed, so I think we can safely answer the parts the OP had trouble with. I've deleted my earlier hints in comments, because a more detailed outline is included here.]
Cars pass checkpoint A in accordance with a Poisson process at an average rate of 24 cars per hour. All cars passing checkpoint A also pass checkpoint B which is located 10 miles down the road. The speed (MPH) of any individual car between A & B is constant and uniformly distributed on (40, 60). The speeds of all cars are mutually independent. Find:
a) $P(40 \text{ cars pass A between noon & 2PM})$
This just uses the fact that the number of cars passing $A$ in two hours is $\text{Poisson}(48)$; it's straight substitution. So abbey was on the right track there.
b) This one is ambiguous, since "between 3 & 6PM, 20 cars pass A at a speed exceeding 55 MPH" might be interpreted either as "of all the cars that pass A in that time period, 20 of them are going faster than 55" or as "exactly 20 cars pass A in the time period, and all of them are going faster than 55". My guess is that the second interpretation is intended.
$P(\text{between 3 & 6PM, 20 cars pass A at a speed exceeding 55 MPH})$
$= P(20 \text{cars pass A in 3 hours})\cdot P(\text{Each car is traveling faster than 55 MPH})$   (independence)
The first term is exactly the same kind of questions as (a), the second is just $(\frac{60-55}{60-40})^20=(\frac{1}{4})^20$; from there it's just calculation.
c) $P(\text{10 cars pass B between noon & 1 PM})$
All cars that pass A eventually pass B, so the average volume of cars of any given speed passing both points is the same.
[Imagine a simpler process where there are only two speeds of car, and two lanes of cars going in one direction, one fast (moving say at 60) and one slow (moving say at 40). A lot of fast cars will pass slow cars, but the typical density of each kind will be unchanged. The fast cars will be passing different slow cars at B than at A, but the density of cars at B is unchanged from that at A.]
As such the process at B looks just like the process at A, and this is another question like (a).
d) the probability that 5 cars pass B between 1 PM & 1:10PM, given that exactly 10 cars pass A between 12:30PM & 1 PM.
In my original hints I suggested drawing a diagram. That's the way to do it, but it's much easier if you set the problem up in a nice way first.
Like so:
Consider a single car.
Let $S$ be the car's speed in miles per minute. $S\sim U(\frac{2}{3},1)$.
Let $W$ be the travel time from $A$ to $B$ in minutes. $W = 10\,S$, so $W\sim U(10,15)$.
Let $T_A$ be the time in minutes after 12:00 that the car passes $A$ (similarly for $T_B$. $T_A\sim U(30,60)$.
$T_A$ and $W$ are independent bivariate uniform. 
Now we draw a diagram, of $W$ and $T_A$, and mark on the region corresponding to $60<T_B<70$:

By inspection, the purple area is 1/4 of the total area. So the probability that a single car which passes A between 12:30 and 1 will pass B between 1 and 1:10 is 1/4.
Cars arrive independently, so P(5 cars pass B between 1 and 1:10 given 10 pass A between 12:30 and 1) is Binomial(10,1/4).
If $X\sim Bin(10,1/4)$ then $P(X=5)= {{10}\choose{5}}.(\frac{1}{4})^5.(\frac{3}{4})^5$
