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.
(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.
Best Answer
You can use inclusion-exclusion when the problem with a more general but opposite constraint is easier. Here you've already computed the solution without constraints. If the constraint would be that the first kid gets at least $6$ oranges, then start giving those oranges, and find $\binom{30-6+9}{30-6}=\binom{33}9$ solutions. But not only could it be another kid that gets too many oranges, there could be more than one kid at once that gets too much. So the "more general opposite" constraint would be, for any set $S$ of kids, that all kids from $S$ get at least $6$ oranges. For this there are $\binom{30-6s+9}{30-6s}=\binom{39-6s}9$ solutions if $S$ has $s$ kids in it.
By inclusion-exclusion you need to count the solutions for $S=\emptyset$, subtract those for $S$ a singleton, add back for $S$ a doubleton, etc. All in all you get $$ \binom{39}9-\binom{10}1\binom{33}9+\binom{10}2\binom{27}9 -\binom{10}3\binom{21}9+\binom{10}4\binom{15}9 -\binom{10}5\binom99 $$ solutions, which if I calculated well gives $2\,930\,456$ possibilities, less than $1.5$% of the original $211\,915\,132$.
Added (much later). Alternatively, one could compute the coefficient of $X^{30}$ in $(1+X+X^2+X^3+X^4+X^5)^{10}=\frac{(1-X^6)^{10}}{(1-X)^{10}}$. This is quite easy (the numerator has only $6$ terms of degree${}\leq30$) and gives the same result, in fact even via the same formula.