Where P(x) is the probability that x number of people do not receive a gift, then
$P(\ge 2) = 1 - (P(0)+P(1))$
There are $10!$ ways that all $10$ people can get one gift and $10^{10}$ ways to distribute $10$ gifts to $10$ people.
$$P(0) = \frac{10!}{10^{10}}$$
Then we have $10$ ways for someone not to get a gift and for each one of those there are $9$ ways for someone to get $2$ gifts and for each of those $2$ people combinations there are $\frac{10!}{2!}$ ways to combine them with $8$ other people who each receive one gift.
$$P(1) = \frac{9\cdot 10\cdot 10!}{10^{10}\cdot 2!}$$
$$P(\ge 2) = 1 - (\frac{10!}{10^{10}}+\frac{9\cdot 10\cdot 10!}{10^{10}\cdot 2!}) = .98330752$$
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
This is a tricky problem. First of all, if you were only looking at one child, the probability that that child did not receive any red chewing gum is
$$\frac{\binom{40}{2}}{\binom{60}{2}}. \tag1 $$
In (1) above, the denominator represents the total number of ways of selecting which two pieces of chewing gum could be presented to one child, while the numerator represents the total number of ways of selecting two pieces of non-red chewing gum for the child.
To attack your problem, I recommend a Combinatorics approach, which is (in effect) what you attempted. I will compute
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}, \tag2 $$
where in (2) above, $D$ represents the total number of ways that the $60$ pieces of chewing gum could be presented to the children, and $N$ represents the total number of ways of doing this, where exactly $(10)$ children do not receive red chewing gum.
For convenience, I will compute
$$D = \binom{60}{2} \times \binom{58}{2} \times \cdots \times \binom{4}{2} \times \binom{2}{2} = \frac{(60)!}{2^{(30)}}. \tag3 $$
This matches your computation. Note that in (2) above, the numerator and denominator must be computed in a consistent manner. For convenience, in (3) above, I regard the order that the children receive the gum as relevant.
That is, I distinguish between child-1 getting two pieces of red gum, child-2 getting two pieces of blue gum, and vice-versa. Therefore, when I compute $N$, I must do so in a consistent manner.
Edit
My computation of $D$ also presumes that for example one piece of red chewing gum is distinguishable from another. This presumption is (also) needed to justify the computation in (3) above.
There are $~\displaystyle \binom{30}{10}~$ ways of selecting exactly which children will not receive any red chewing gum.
Assume that the children who did not receive red chewing gum are child-1, child-2, ..., child-10.
The number of ways that this can occur is
$$\binom{40}{2} \times \binom{38}{2} \times \cdots \times \binom{24}{2} \times \binom{22}{2} = \frac{(40)!}{2^{(10)} \times [(20)!]}. \tag4 $$
Under the above hypothetical, child-11, child-12, ..., child-30, have not yet received any gum. Also, at this point, there are $40$ pieces of gum left, of which exactly $(20)$ are red.
Edit
A trap here, is that for this hypothetical to work, each of the $(20)$ remaining children must each receive at least one piece of red chewing gum. Otherwise, the number of children who did not receive red chewing gum, in this hypothetical would not be exactly equal to $(10)$.
So, you have $(20)$ children left, $(20)$ pieces of red chewing gum left, and $(20)$ pieces of non-red chewing gum left. The number of ways that these $(40)$ pieces of chewing gum can be distributed so that each of these $(20)$ children receives $(1)$ red piece of gum and $(1)$ non-red piece of gum is
$$[(20)!] \times [(20)!]. \tag5 $$
That is, you imagine that the child-11, child-12, ..., child-30 are lined up, and that the red gum and non-red gum will also be lined up to hand to each child.
Final computation.
$$N = \binom{30}{10} \times \frac{(40)!}{2^{(10)} \times [(20)!]} \times [(20)!]^2$$
$$ = \frac{(30)!}{(10)! \times [(20)!]} \times \frac{(40)!}{2^{(10)} \times [(20)!]} \times [(20)!]^2 $$
$$ = \frac{[(30)!] \times [(40)!]}{[(10)!] \times 2^{(10)}}.$$
$$D = \frac{(60)!}{2^{(30)}}. $$
Therefore,
$$\frac{N}{D} = \frac{[(30)!] \times [(40)!]}{[(10)!] \times 2^{(10)}} \times \frac{2^{(30)}}{(60)!}$$
$$ = \frac{[(30)!] \times [(40)!] \times 2^{(20)}}{[(10)!] \times [(60)!]}.$$