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
$${16\over19}\cdot20!=16\cdot20\cdot18!$$
You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.
Best Answer
Hint: Imagine that you have ten chairs. Set out six empty chairs in a row, leaving spaces between them and at the ends of the row in which to place the four people.
$$\square c \square c \square c \square c \square c \square c \square$$
Choose four of those seven spaces in which to place the chairs of the four people, then arrange the people in those chosen spaces.