Your answers to the first two questions are correct.
In how many ways can $15$ indistinguishable candies be distributed to five children if child $C$ and child $D$ receive $7$ candies together?
We must distribute seven candies among the children $C$ and $D$ and eight candies among the children $A$, $B$, and $E$. The number of ways we can distribute the candies to children $C$ and $D$ is eight since $C$ must receive between $0$ and $7$ candies inclusive, with $D$ receiving the rest. The number of ways the remaining eight candies can be distributed to the children $A$, $B$, and $D$ is
$$\binom{8 + 3 - 1}{3 - 1} = \binom{10}{2}$$
as you correctly found. Hence, the number of ways of distributing to the five children if $C$ and $D$ receive exactly seven candies between them is
$$\binom{7 + 2 - 1}{2 - 1}\binom{8 + 3 - 1}{3 - 1} = \binom{8}{1}\binom{10}{2}$$
In how many ways can $15$ indistinguishable candies be distributed to five children if no child receives more than six candies?
Let $x_i$, $1 \leq i \leq 5$, be the number of candies received by the $i$th child. Then we seek the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15 \tag{1}$$
in the nonnegative integers subject to the restrictions that $x_i \leq 6$ for $1 \leq i \leq 5$.
A particular of equation 1 corresponds to the placement of four addition signs in a row of $15$ ones. For instance,
$$1 1 1 + 1 1 1 1 + 1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, $x_3 = 2$, $x_4 = 0$, and $x_5 = 6$. The number of such solutions is the number of ways we can place four addition signs in a row of fifteen ones, which is
$$\binom{15 + 5 - 1}{5 - 1} = \binom{19}{4}$$
since we must choose which four of the nineteen positions required for fifteen ones and four addition signs will be filled with addition signs.
By similar reasoning, the number of solutions of the equation
$$x_1 + x_2 + x_3 + \cdots + x_n = k$$
in the nonnegative integers is
$$\binom{k + n - 1}{n - 1}$$
since we must choose which $n - 1$ of the $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs.
From these, we must subtract those cases in which at least one child receives more than six candies. Observe that at most two children could receive more than six candies since $2 \cdot 7 = 14 < 15 < 21 = 3 \cdot 7$.
Suppose a child receives more than six candies. There are five ways to choose that child. We give that child seven candies. The remaining eight candies can be distributed among the five children in
$$\binom{8 + 5 - 1}{5 - 1} = \binom{12}{4}$$
ways. Hence, there are
$$\binom{5}{1}\binom{12}{4}$$
ways to distribute the candies in such a way that a child receives more than six candies.
However, if we subtract this amount from the total, we will have subtracted too much since we have counted each case in which two children receive more than six candies twice, once for each way of designating one of those children as the child who received more than six candies. We only want to subtract those cases once, so we must add those cases back.
Suppose two children each receive more than six candies. There are $\binom{5}{2}$ ways to select those two children. Give each of them seven candies. That leaves one candy to distribute among the five children, which can be done in five ways. Hence, the number of distributions in which two children receive more than six candies is
$$\binom{5}{2}\binom{1 + 5 - 1}{5 - 1} = \binom{5}{2}\binom{5}{4}$$
By the Inclusion-Exclusion Principle, the number of ways of distributing $15$ indistinguishable candies to five children so that no child receives more than six candies is
$$\binom{19}{4} - \binom{5}{1}\binom{12}{4} + \binom{5}{2}\binom{5}{4}$$
In how many ways can $15$ indistinguishable candies be distributed to five children if each child receives a different amount of candies.
List the ways $15$ can be expressed as a sum of five distinct nonnegative numbers, starting with
$$0 + 1 + 2 + 3 + 9 = 15$$
and ending with
$$1 + 2 + 3 + 4 + 5 = 15$$
For each of these ways (there are not many), there are $5!$ ways to distribute the candies to the children, depending on which child receives which number of candies.
Your answers to the first two questions are correct, as is your second method for the third problem.
Your first method for the third problem over counts. To see this, suppose the candy bars the children receive are $A, B, C, D, E, F, G, H, I$. You count each distribution multiple times. For instance, if we give $A, B$ to the first child, $C, D, E$ to the second child, and $F, G, H, I$ to the third child, your method counts this case $2 \cdot 3 \cdot 4 = 24$ times, once for each of the two ways of designating one of the candy bars the first child receives as the candy bar reserved for that child, once for each of the three ways of designating one of the three candy bars received by the second child as the candy bar reserved for that child, and once for each of the four ways of designating one of the four candy bars the third child receives as the one reserved for that child. To illustrate a few ways you count this distribution, consider the following table.
\begin{array}{c | c |c | c | c | c}
R_1 & R_2 & R_3 & A_1 & A_2 & A_3\\ \hline
A & C & F & B & D, E & G, H, I\\
B & C & F & A & D, E & G, H, I\\
A & D & F & B & C, E & G, H, I\\
B & D & F & A & C, E & G, H, I\\
A & E & F & B & C, D & G, H, I\\
B & E & F & A & C, D & G, H, I
\end{array}
where $R_i$, $1 \leq i \leq 3$, denotes the candy bar reserved for the $i$th child, and $A_i$, $1 \leq i \leq 3$, represents the additional candy bars received by the $i$th child.
Best Answer
We can consider two cases: one child receives three candies and the others each receive one, or two children each receive two candies and the other child receives one.
One child receives three candies and the others each receive one: There are three ways to select the select who receives three candies, $\binom{5}{3}$ ways to select which three of the five candies that child receives, and $2!$ ways to distribute the remaining candies to the remaining children so each of them receives one candy. Hence, there are $$\binom{3}{1}\binom{5}{3}2! = 60$$ such distributions, as you found in the comments.
Two children each receive two candies and the other child receives one: There are $\binom{3}{2}$ ways to select the children who will receive two candies. There are $\binom{5}{2}$ ways to give two of the five candies to the younger of those two children and $\binom{3}{2}$ ways to give two of the remaining three candies to the older of those two children. The remaining child must receive the remaining candy. Hence, there are $$\binom{3}{2}\binom{5}{2}\binom{3}{2} = 90$$ such distributions.
Total: Since the above cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{5}{3}2! + \binom{3}{2}\binom{5}{2}\binom{3}{2} = 150$$ ways to distribute five candies to three children so that each child receives at least one candy.
This problem can also be solved using the Inclusion-Exclusion Principle. There are three possible recipients for each candy, so there are $3^5$ possible distributions.
From these, we subtract those in which a child does not receive a candy. There are three ways to exclude one of the children from receiving a candy and $2^5$ ways to distribute the candies to the remaining children. There are $3 \cdot 2^5$ such cases.
However, if we subtract $3 \cdot 2^5$ from the total, we will have subtracted too much since we will have subtracted the three cases in which all the candies are given to one child twice, once for each way of designating one of the children who has not received a candy as the child who does not receive a candy.
Therefore, we must add those three cases to our running total, which yields $$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 150$$ ways to distribute five candies to three children so that each child receives at least one candy.