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.
In the stars-and-bars formulation of the second problem, there are $17$ stars (cakes), but only $5$ bars (children) to place in the $17-1=16$ spaces between adjacent stars, because those $5$ bars define $6$ partitions around them (before, between and after) which are identified with the $6$ children.
Then the number of ways to assign cakes to children is the same as the number of ways to choose, among the $16$ spaces, $5$ spaces to hold the bars, hence $\binom{11+6-1}5$ and not $\binom{11+6-1}6$.
Best Answer
Your first solution is correct. There are three possible recipients for each of the six candies, so there are $3^6$ possible distributions of the candies.
If you wish to consider how many candies each child receives, you need to consider cases based on partitions of $6$ into at most three parts. \begin{align*} 6 & = 6\\ & = 5 + 1\\ & = 4 + 2\\ & = 4 + 1 + 1\\ & = 3 + 3\\ & = 3 + 2 + 1\\ & = 2 + 2 + 2 \end{align*}
One child receives six candies: There are three ways to select the recipient of all six candies.
One child receives five candies and another child receives one candy: There are three ways to select the child who will receive five candies, $\binom{6}{5}$ ways to select which five of the six candies that child receives, and two ways to choose the child who receives the remaining candy. Hence, there are $$\binom{3}{1}\binom{6}{5}\binom{2}{1}$$ such distributions.
One child receives four candies and another child receives two candies: There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose the child who receives the remaining two candies. Hence, there are $$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$ such distributions.
One child receives four candies and each of the other children receive one candy each: There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose which of the remaining two candies the younger of the two remaining children receives. The other child must receive the remaining candy. Hence, there are $$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$ such distributions.
Two children each receive three candies: There are $\binom{3}{2}$ ways to select which two children receive three candies each and $\binom{6}{3}$ ways to select which three of the six candies the younger of the two selected children will receive. The other selected child must receive the three remaining candies. Hence, there are $$\binom{3}{2}\binom{6}{3}$$ such distributions.
One child receives three candies, a second child receives two candies, and the third child receives one candy: There are three ways to select the child who will receive three candies, $\binom{6}{3}$ ways to select which three candies that child will receive, two ways to decide which of the remaining children will receive two candies, and $\binom{3}{2}$ ways to select which two of the remaining three candies that child will receive. The remaining child must receive the remaining candy. Hence, there are $$\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}$$ such distributions.
Each of the three children receives two candies: There are $\binom{6}{2}$ ways to select which two candies are received by the youngest child and $\binom{4}{2}$ ways to select which two of the remaining four candies are received by the next youngest child. The oldest child must receive the two remaining candies. Hence, there are $$\binom{6}{2}\binom{4}{2}$$ such distributions.
Total: Since the above cases are mutually exclusive and exhaustive, the number of ways of distributing six different candies to three children is $$\binom{3}{1} + \binom{3}{1}\binom{6}{5}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{2}\binom{6}{3} + \binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2} + \binom{6}{2}\binom{4}{2} = 3^6$$