I'm re-writing your conditions slightly to be clear about each case.
(a) $n$ different candies into $t$ different boxes:
Since each candy can be placed into one of $t$ different boxes, the number of ways will be
$$t \cdot t \cdot t\cdot t \cdots t = t^n$$
(b) $n$ identical candies into $t$ different boxes, no box to remain empty:
Your solution is correct.
(c) $n$ identical candies into $t$ different boxes, some boxes may remain empty:
[I'm taking "candies equally new" to mean they're identical; if that's not the case please specify what "equally new" means.]
Since the candies are identical, there's no need to choose a candy for each box as you're doing in the early part of your solution. The simplest way to do this is similar to your solution for (b), except now some boxes may remain empty; thus we need the number of arrangements of $(t-1)$ partitions and $n$ candies, i.e.
$$\frac{(n+t-1)!}{n!(t-1)!} = \binom{n+t-1}{n} \quad \text{or} \quad \binom{n+t-1}{t-1} $$
(d) & (e) $n$ different candies into $t$ identical boxes
The solution(s) here are known as Stirling Numbers of the Second Kind which have no closed formula but do satisfy a recurrence relationship.
For (d) if all the boxes are to be unempty, then the number of ways is $$S(n,t)$$ For (e) if some boxes may remain empty, then the number of ways is $$S(n,1) + S(n,2) + S(n,3) + \cdot + S(n,t) = \sum_{k=1}^t S(n,k)$$.
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
$$\binom{3}{1} + \binom{3}{1}\binom{6}{5}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{2}\binom{6}{3} + \binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2} + \binom{6}{2}\binom{4}{2} = 3^6$$
Best Answer
You have 35 candies to choose. Let's say you have four "bars"; they separate the types of candies from each other. e.g.:
-------|-------|-------|-------|-------
Means 7 candies of each. The answer is 39C4 because this is only a rearrangement problem.
The second one, assume that each has one candies. Then it's just 30 candies to choose and 4 bars again.
Same for the last one; you choose 35-2-4=29 candies with 4 bars.
If you want to learn more, search "sticks and bars". It's a powerful method.