In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue?
You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue.
Method 1: We treat the block of five girls as a single object. We then have six objects to arrange, the block of girls and the five boys. The objects can be arranged in $6!$ ways. The five girls can be arranged within the block in $5!$ ways. Thus, there are $6!5!$ ways for five boys and five girls to stand in a queue if all five girls stand consecutively in the queue.
Method 2: Line up the five boys, which can be done in $5!$ ways. This creates six spaces in which to place the block of five girls, four between successive boys and two at the ends of the row.
$$\square b_1 \square b_2 \square b_3 \square b_4 \square b_5 \square$$
Choose one of these six spaces in which to place the block of girls, then arrange the five girls within the block. This can be done in $6 \cdot 5!$ ways. Hence, the number of admissible arrangements is $6!5!$.
In how many ways can $5$ boys and $5$ girls stand in a queue if exactly four girls stand consecutively in the queue?
We modify the second method above.
Line up the five boys in $5!$ ways. This creates six spaces in which to place the girls. Choose which four of the five girls stand consecutively, which can be done in $\binom{5}{4}$ ways. Choose which of the six spaces the block of four girls fills. Arrange the four girls in that space in $4!$ ways. That leaves five spaces in which to place the remaining girl. Hence, the number of ways five boys and five girls can stand in a queue if exactly four girls stand consecutively is
$$5!\binom{5}{4} 6 \cdot 4! \cdot 5 = 5! \cdot 5 \cdot 6 \cdot 5 \cdot 4! = 5 \cdot 6!5!$$
In how many ways can $20$ be expressed as the sum of five distinct increasing positive integers?
Since $20$ is a small number, we can simply write down all the possibilities:
\begin{align*}
20 & = 1 + 2 + 3 + 4 + 10\\
& = 1 + 2 + 3 + 5 + 9\\
& = 1 + 2 + 3 + 6 + 8\\
& = 1 + 2 + 4 + 5 + 8\\
& = 1 + 2 + 4 + 6 + 7\\
& = 1 + 3 + 4 + 5 + 7\\
& = 2 + 3 + 4 + 5 + 6
\end{align*}
Notice that any sum of five distinct positive integers is at least $1 + 2 + 3 + 4 + 5 = 15$. We then have to distribute five more ones in such a way that we preserve the increasing sequence. Since $5$ can be partitioned into at most five positive integers in the following seven ways,
\begin{align*}
5 & = 5\\
& = 4 + 1\\
& = 3 + 2\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1\\
& = 2 + 1 + 1 + 1\\
& = 1 + 1 + 1 + 1 + 1
\end{align*}
we can do so in the following ways:
\begin{align*}
(0, 0, 0, 0, 5)\\
(0, 0, 0, 1, 4)\\
(0, 0, 0, 2, 3)\\
(0, 0, 1, 1, 3)\\
(0, 0, 1, 2, 2)\\
(0, 1, 1, 1, 2)\\
(1, 1, 1, 1, 1)\\
\end{align*}
Adding these, respectively, to the vector $(1, 2, 3, 4, 5)$ yields the solutions
\begin{align*}
(1, 2, 3, 4, 10)\\
(1, 2, 3, 5, 9)\\
(1, 2, 3, 6, 8)\\
(1, 2, 4, 5, 8)\\
(1, 2, 4, 6, 7)\\
(1, 3, 4, 5, 7)\\
(2, 3, 4, 5, 6)
\end{align*}
that correspond to the seven sums we wrote above.