Your introduction of the factor ${4\choose2}2!2!$ specifies who sits where in only one car. You need to raise that factor to the fifth power, to account for all five cars.
Note, though, that ${4\choose2}2!2!={4!\over2!2!}2!2!=4!$, so the fifth power will entirely cancel the $(4!)^5$ in the $20!\over(4!)^5$, leaving just $20!$. This makes sense: If it matter exactly where people sit in the cars, then the number of arrangements amounts to assigning a different seat to each of the $20$ people, which is just $20!$.
Added in response to OP's comment: As to the second part, one way, as you note, is to count the number of ways to have the two people sit in the case car, which is $20\cdot3\cdot18!$, and subtract from $20!$. But another way is to have those two people -- let's call them Bob and Ted -- be the first two to take seats. If we seat Bob first, he has $20$ choices, but then Ted has just $16$. After that, it's $18\cdot17\cdots2\cdot1=18!$ ways to seat everyone else, for a total of $20\cdot16\cdot18!$ different seatings with Bob and Ted in different cars.
Note, this generalizes easily to more people who don't want to sit in the same car, say Bob and Carol and Ted and Alice: $20\cdot16\cdot12\cdot8\cdot16!$, whereas the subtraction approach gets messy.
Added yet later: There's one more approach that's worth mentioning. Have Ted be the very last person to sit. It's easy to see that Bob and the others have $20\cdot19\cdots3\cdot2=20!$ ways of sitting. No matter where Bob sits, the proability that the empty seat is in another car is $16\over19$, so the number of ways to keep Bob and Ted apart is
$${16\over19}\cdot20!=16\cdot20\cdot18!$$
Let $x$ denote the number of cars and $y$ the number of motorcycles in a configuration. Then we have the condition:
$$
2x+y=n
$$
where $x,y\geq 0$. This has solutions $x=0,1,...,\lfloor n/2\rfloor$ and $y=n-2x$. For each such solution there will be
$$
\binom{x+y}{x}=\frac{(x+y)!}{x!y!}
$$
ways to distribute the sequence of $x$ cars among $x+y$ vehicles. Hence the figure becomes
$$
P(n)=\sum_{x=0}^{\lfloor n/2\rfloor}\binom{n-x}{x}
$$
since $y=n-2x$ implies $x+y=n-x$. Perhaps this can be expressed in some different form which is more neat.
Heureka! It is indeed the Fibonacci sequence:
$$
P(n)=P(n-1)+P(n-2)
$$
Since you can form all new configurations ending in a motorcycle by adding a motorcycle to the end of one of the $(n-1)$-configs, and similarly all ending in a car by adding a car to the end of one of the $(n-2)$-configs. Those two are necessarily distinct because they end in a different kind of vehicle. Hence we just add them together.
Best Answer
(a) Good work, your answer is correct.
(b) There are two possible interpretations of the problem; as you indicate, it's unclear whether or not the point of beginning matters, or in other words, whether any rotation of the cars around circle is the same.
If the point of beginning does not matter, your answer of $3!$ is exactly right.
If the point of beginning does matter, you would proceed similarly to (a), except there are $10$ possible points of beginning instead of $7$.