$1)$ There are $4^3$ ways to distribute the teddy bears and $\dbinom{9+4-1}{9}$ ways of distributing the lollipops.
In total there are $4^3\times\dbinom{9+4-1}{9}$ ways without restriction.
$2)$ Since each can have at most one teddy bear, so one child must go without. The teddy bears can be distributed to $3$ children in $3!$ ways. We can distribute the lollipops in $\dbinom{9+4-1}{9}$ ways. There are $4$ ways to choose a child who will not receive a teddy bear. In total there are $4\times3!\times\dbinom{9+4-1}{9}$ ways.
$3)$ We can distribute $3$ teddy bears to $4$ children in $4^3$ ways.
We can ensure that each child gets $3$ goodies by filling out each child with lollipops.
So, there are a total of $4^3=64$ ways
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.
Best Answer
(b) To distribute 15 candies to five children there are $5^{15}$ ways. To distribute the same candies to only four children, you have to choose which child will definitely not get any candy, which is $\binom51$ ways, and then actually distribute ($4^{15}$ ways). We continue down this line: distributing to three children is pre-omitting two children in $\binom52$ ways and distributing in $3^{15}$ ways, etc.
Thus the final total is the number of ways to distribute to five children, minus the ways to distribute to four, plus the ways to distribute to three, minus the ways to distribute to two, plus the ways to distribute to one: $$5^{15}-5\cdot4^{15}+10\cdot3^{15}-10\cdot2^{15}+5\cdot1^{15}$$
(d) Suppose A, B get exactly four candies. There are $\binom{15}4$ ways to choose the candies that go to them, $2^4$ ways to distribute that special selection and $3^{11}$ ways to distribute the other 11 candies to C, D, E.
Repeat for A, B receiving 3, 2, 1, 0 candies and add up: $$\binom{15}42^43^{11}+\binom{15}32^33^{12}+\binom{15}22^23^{13}+\binom{15}12^13^{14}+\binom{15}02^03^{15}$$
(e) First you choose which two children get nothing, which is $\binom52$ ways. The other three children must get at least one candy, so we can carry the argument in the answer to (b) over (include giving to three, exclude giving to two, include giving to one): $$\binom52\left(3^{15}-3\cdot2^{15}+3\cdot1^{15}\right)$$
Lastly, (a) and (c) are correct. In particular, (c) can be derived in a simpler way: have the children receive their candies in a queue, three at a time ($15!$ ways), then divide by $6^5$ for the $3!$ ways each child can permute their sweets without affecting their selection, thus $\frac{15!}{6^5}$ ways.