(1) We will first solve a different problem: How many ways are there to split the group into two teams of $4$, one team to wear blue uniforms, and the other to wear red. That's easy, there are $\binom{8}{4}$ ways to pick the people who will wear blue.
Now we solve our problem. Every one of the $N$ ways to split the group into two teams gives rise to $2$ ways to split them into uniformed teams. So $2N=\binom{8}{4}$ and now we know $N$.
(2) We count the ways to split the group into two uniformed teams, each with $2$ girls and $2$ boys. There are $\binom{4}{2}$ ways to pick the girls who will wear blue, and for each such way there are $\binom{4}{2}$ ways to pick the boys who will wear blue, for a total of $\binom{4}{2}\binom{4}{2}$.
Now the reasoning of (1) shows that the number of ways to split into two groups, each with $2$ boys and $2$ girls, is $\frac{1}{2}\binom{4}{2}\binom{4}{2}$.
(3) By the reasoning of (1), there are $\frac{1}{2}\binom{8}{4}$ ways to split the group into two teams of $4$. There is only $1$ way to split them so that one team is all girls and the other all boys. So there are $\frac{1}{2}\binom{8}{4}-1$ ways to split the group into two teams, each of which has at least $1$ girl and at least $1$ boy.
Remark: The problem is somewhat ambiguous. Another interpretation is that we want to split the group of $8$ into two groups, not necessarily equal in size (except for question (2), where equality is forced). It is not hard to answer (1), but (3) for this interpretation is a bit messy, division into cases is best.
