Method 1: We can arrange the eight girls in a row in $8!$ ways. This creates nine spaces in which to insert the boys, seven between successive girls and two at the ends of the row.
$$\square g \square g \square g \square g \square g \square g \square g \square g \square$$
To ensure that no two of the boys sit in consecutive seats, we choose six of these nine spaces in which to insert the boys, which we can do in $\binom{9}{6}$ ways. The boys can be arranged in the six selected spaces in $6!$ orders, so the number of seating arrangements of eight girls and six boys in which no two of the boys sit in consecutive seats is $$8!\binom{9}{6}6!$$ as you found.
Method 2: This is a modification of your approach.
We can arrange the six boys in a row in $6!$ ways. This creates seven spaces in which to place the girls, five between successive boys and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$
If $x_k$ represents the number of girls placed in the $k$th space, then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 8 \tag{1}$$
The requirement that no two boys sit in adjacent seats means that $x_2, x_3, x_4, x_5, x_6 \geq 1$, while $x_1, x_7 \geq 0$. If we let $y_k = x_k - 1$, $2 \leq k \leq 6$, then $y_k$ is a nonnegative integer. Substituting $y_k + 1$ for $x_k$, $2 \leq k \leq 6$, in equation 1 yields
\begin{align*}
x_1 + y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 + y_5 + 1 + y_6 + 1 + x_7 & = 8\\
x_1 + y_1 + y_2 + y_3 + y_4 + y_5 + y_6 + x_7 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution corresponds to the insertion of six addition signs in a row of three ones. The number of such solutions is
$$\binom{6 + 3}{6} = \binom{9}{6}$$
since we must choose which three of the nine positions (three ones and six addition signs) will be filled with additions signs. Finally, we can arrange the eight girls in the selected positions in $8!$ ways, again yielding $$6!\binom{9}{3}8!$$
Your second answer is correct.
In how many ways can six boys and four girls be seated around a table if no two girls are adjacent?
Method 1: Suppose one of the boys is Asa. We will use him as a reference point. The remaining five boys can be seated in $5!$ ways as we proceed clockwise around the table from Asa. Seating the six boys creates six spaces in which a girl could be placed, one to the left of each boy. To separate the girls, we must choose four of these six spaces in which to place a girl. The girls can be arranged in the four selected spaces in $4!$ ways as we proceed clockwise around the table from Asa. Hence, the number of admissible seating arrangements is
$$5!\binom{6}{4}4!$$
Method 2: We modify your attempt. Suppose Adrienne is one of the girls. The other girls can be seated in $3!$ ways as we proceed clockwise around the table from Adrienne. This creates four spaces in which to place the boys, one to the left of each girl. Let $x_1, x_2, x_3, x_4$ denote, respectively, the number of boys in the first, second, third, and fourth spaces as we proceed clockwise around the table from Adrienne. Since there are a total of six boys,
$$x_1 + x_2 + x_3 + x_4 = 6$$
Since no two of the girls are adjacent, there must be at least one boy in each of the four spaces. Hence, this is an equation in the positive integers. A particular solution of the equation corresponds to placing three addition signs in the five spaces between successive ones in a row of six ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, placing an addition sign in the first, second, and fourth spaces gives
$$1 + 1 + 1 1 + 1 1$$
which corresponds to the solution $x_1 = 1$, $x_2 = 1$, $x_3 = 2$, $x_4 = 2$. The number of such solutions is the number of ways we can place three addition signs in the five spaces between successive one in a row of six ones, which is
$$\binom{6 - 1}{4 - 1} = \binom{5}{3}$$
One the number of boys in each space has been selected, the boys can be arranged in those spaces in $6!$ ways as we proceed clockwise around the table from Adrienne. Hence, there are
$$3!\binom{5}{3}6!$$
admissible seating arrangements.
Method 3: We correct your approach. Again, suppose Adrienne is one of the girls. The other girls can be seated in $3!$ ways as we proceed clockwise around the table from Adrienne. Choose which four of the six boys will sit to the immediate left of a girl. Those boys can be arranged in $4!$ ways as we proceed clockwise around the table from Adrienne. That leaves two boys to place. There are two possibilities: both boys are placed between the same two girls so that there are three boys between those girls or they are placed between separate pairs of girls so that there are two pairs of girls with two boys between them.
Both of the remaining boys are placed between the same two girls: There are four ways to choose the pair of girls the boys sit between. Both boys must sit to the left of the boy who has already been seated to the immediate left of the first girl we encounter in the pair as we proceed clockwise around the table from Adrienne (starting with Adrienne). There are two ways to choose which of the boys who has not yet been seated sits next to that boy. Hence, there are $\binom{4}{1}2!$ such cases.
The remaining boys are placed between two different pairs of girls: There are $\binom{4}{2}$ ways to choose which two pairs of girls the boys sit between. In each case, the boy must be seated to the left of the boy who has already been seated to the immediate left of the first girl we encounter in the pair as we proceed clockwise around the table from Adrienne (starting with Adrienne). There are two ways to choose which of the boys we will encounter first as we proceed clockwise around the table from Adrienne. There are $\binom{4}{2}2!$ such cases.
Thus, there are
$$3!\binom{6}{4}4!\left[\binom{4}{1}2! + \binom{4}{2}2!\right]$$
admissible seating arrangements.
Your errors:
You forgot to arrange the four boys you placed in the gaps, so you were missing a factor of $4!$. Had you included that factor, your count would have been too large. The reason for this is that you counted the same arrangement more than once. For instance, if Asa, Bradley, and Charles all sit between Adrienne and Bronwyn, your approach would count the same arrangement $3! = 6$ times, corresponding to the $3!$ orders in which the same boys could be placed in the same seats. If Asa and Bradley were to be placed between Adrienne and Bronwyn and Charles and David were to be placed between Bronwyn and Claire, then your approach would count the same arrangement four times, corresponding to the $2!$ orders in which Asa and Bradley could be placed in the same seats and the $2!$ orders in which Charles and David could be placed in the same seats.
Best Answer
I think your second answer is not correct.
For the first it's just $$\frac{8!}{8}=7!$$ because there are $8!$ permutations of $8$ children and since all this happens in the circle we need to divide by $8$.
The second problem: $$4!4!$$ Because let $a$ be all four girls and $b$ be all four boys.
Thus, we need to place $a$ and $b$ on the circle, which gives $\frac{2!}{2}=1$ cases.
Now, we can remember that they are $4$ (boys and girls), which gives $(4!)^2$.
The third problem: $$3!4!$$
For the new second problem.
If you mean that we need $4$ boys placed together or $4$ girls placed together then the answer is $$2\cdot\frac{5!4!}{5}$$