A simple way is to suppose that person A has been assigned to Group 1. Note that $3$ of the remaining $11$ will be assigned to Group 1. The probability that person B is one of them is $\frac{3}{11}$. So the probability that A and B end up in different groups is $\frac{8}{11}$.
There are more elaborate combinatorial arguments.
Remark: At the request of OP, here is a more complicated argument. Without changing the probabilities, we may assume that the groups are named groups, say Groups 1, 2, and 3.
The number of (equally likely) ways to assign people to named groups is $\frac{12!}{4!4!4!}$. One way of seeing this is that there are $\binom{12}{4}$ ways to decide who goes into Group 1, and for each of these there are $\binom{8}{4}$ ways to decide who goes into Group 2.
Now we could directly count the number of ways to assign A and B to different groups, or do it indirectly by counting the number of ways to assign A and B to the same group. We do the second.
There are $\binom{3}{1}$ ways to choose the common group. For each of these, there are $\binom{10}{2}$ ways to choose the groupmates of A and B. And there are $\binom{8}{4}$ ways to decide on the members of the first unused group, for a total of $\binom{3}{1}\binom{10}{2}\binom{8}{4}$.
Divide by $\frac{12!}{(4!)^3}$ to find the probability A and B end up in the same group. Fairly quickly, the expression simplifies to $\frac{3}{11}$.
Alternatively, there are $\binom{8}{4}$ ways to chose the positions of the girls, and only $2$ ways of picking the locations so that they alternate with the boys, for a probability of $$\frac{2}{\binom{8}4}$$
Best Answer
A much simpler approach is to note that one group has $2N$ children. How many ways to select $2N$ children out of $4N?$ Then among the groups of $2N$ children, how many have $N$ boys in them?