You have solved the problem by first counting the arrangements you don't want. This is a perfectly valid idea and often a simple way to do these problems, however I think in this case the direct approach will be easier.
First note that neither B nor G can occupy the centre square. So
- Choose a square for B. . . . . $6$ ways.
- Choose a square for G. . . . . $3$ ways
- Place the other $5$ blocks. . . . .$5!$ ways.
Answer, $3\times6\times5!=2160$.
Taking your approach, it seems to me that you are attempting to count things in the right way, but your actual numbers are wrong. It should be $7!-(6\times2\times5!+2\times6\times5!)$ which is also $2160$.
Let's look at the combinations, and we'll do it one group at a time. We are looking for the probability that each of the four groups has exactly one boy and three girls.
For the first group, we are choosing $4$ people out of $16$. One is a boy, out of $4$ possibilities. Three are girls, out of $12$ possibilities. The choices of boy and girls are independent, so we can multiply their counts. Therefore, the probability that the first group has one boy and three girls is
$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}}=\frac{44}{91}$$
Now for the second group. After the first group is chosen suitably, we have to choose $1$ boy out of $3$ remaining and $3$ girls out of $9$ remaining. But the total number of possible groups is $4$ people out of $12$ remaining. Therefore, given that the first group was successful, the probability that the second group is successfully chosen is
$$\frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}}=\frac{28}{55}$$
Now for the third group. After the first two groups are chosen suitably, we have to choose $1$ boy out of $2$ remaining and $3$ girls out of $6$ remaining. But the total number of possible groups is $4$ people out of $8$ remaining. Therefore, given that the first and second groups were successful, the probability that the third group is successfully chosen is
$$\frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}}=\frac{4}{7}$$
If the first three groups are correct, so is the fourth. Therefore, the total probability of success with all four groups is
$$\frac{44}{91}\cdot\frac{28}{55}\cdot\frac{4}{7}=\frac{64}{455}$$
Or, if you want the final probability as one large calculation,
$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}} \cdot
\frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}} \cdot
\frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}}
=\frac{64}{455}$$
Best Answer
For (1), note that the objects are not all distinct: e.g. the green blocks. consider the green blocks in isolation. We can arrange them in $4!$ ways if they were distinct- however, as they are all the same colour, then we need to divide by a factor of $4!$, giving a total arrangements of 1. This is because we can swap the green blocks around with each other any way we like and still get the same arrangement.
Now, if all the blocks of the same colours are together- all the greens are together, all the blues are together and all the reds are together. The greens can be arranged amongst themselves in only 1 way, as I showed; similar logic goes for the red and blue blocks. So now we have 3 objects to arrange: the blob of green blocks, red blocks and blue blocks. They are all distinct objects, so $3!=6$ ways.
As for (2)- let's consider the digits 3, 8 and 5 in isolation. There are 3!=6 arrangements of these 3 digits. Only one arrangement has this particular order: 3,8,5. So it follows that only $\frac{1}{6}$ of the total arrangements will satisfy the given condition.
Hence, for (2) the answer is $\frac{8!}{6}$