We make a few comments. 1(a)(i) is right. For 1(a)(ii), I would choose the two kids who get $7$ toys each. That can be done in $\binom{4}{2}$ ways. The toys for the shorter of these two kids can be chosen in $\binom{18}{7}$ ways, then the toys for the taller one can be chosen in $\binom{11}{7}$ ways, then the toys for the shorter of the two remaining kids can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}\binom{18}{11}\binom{11}{7}\binom{4}{2}$.
The question 1(b) uses standard Stars and Bars. 1(b)(i) is attacked correctly. But 1(b)(ii) is quite wrong, The number of ways to distribute the pears so that each gets at least one is $\binom{6}{2}$.
Question 2 is right apart from notational issues, you should write ${}_{30}P_{20}$, for example.
Remark: It is not a good idea to ask a many parts question on MSE, particularly if the parts are not closely related.
Suppose that the couples are Adam and Alice, Bob and Brenda, and Cathy and Charles.
If you just want the answer, you can calculate it online with Sage.
Open a notebook and type
D = Derangements(['A','A','B','B','C','C'])
D.cardinality()
and click "evaluate". The answer is 10.
You can, of course, choose different numbers of couples. You could also throw in singles, triples, etc.
If you want a list of them, then evaluate
D.list()
For instance, the entries in ['B','B','C','C','A','A'] are the first letter in the names drawn by Adam, Alice, Bob, Brenda, Cathy, and Charles respectively.
There are 8 ways to assign names to each arrangement, so the number of named permutations is 80. This result also follows from the formula in joriki's answer here. We have
$$\int_0^\infty (2-4x+x^2)^3 e^{-x}\,dx=80. $$
Update:
The Sage program that I suggested outputs a list of ten permutations, for example: ['B', 'B', 'C', 'C', 'A', 'A'].
What does this mean, and how come the answer to this question is sometimes 10 and sometimes 80?
First of all, we may as well suppose that the six people line up in alphabetical order to randomly choose a name.
We interpret ['B', 'B', 'C', 'C', 'A', 'A'] as follows:
$$\begin{array}{c||c|c|c|c|c|c}
\text{Chooser} &\text{Adam}&\text{Alice}&\text{Bob}&\text{Brenda}&\text{Cathy}&\text{Charles}\\ \hline
\text{First letter in}&\text{B}&\text{B}&\text{C}&\text{C}&\text{A}&\text{A}\\
\text{name chosen}\\
\end{array}$$
There are a total of ${6!\over 2!\,2!\,2!}=90$ possible permutations, and exactly 10 of them (including the one above)
correspond to the situation where nobody gets their own name or the name of their partner.
This is why the Sage program returns the answer 10.
Notice that the table above does not tell us what person chose which name, as
this information is not needed. All we need to know is that everyone chose a name
whose first letter differs from the first letter of their own name. The good permutations are
those with no $A$s in the first two spots, no $B$s in the second two spots, and no $C$s in the final two spots.
Suppose, though, that we want all that information anyway. One possible
outcome consistent with the example above is:
$$
\begin{array}{c||c|c|c|c|c|c}
\text{Chooser} &\text{Adam}&\text{Alice}& \text{Bob}& \text{Brenda}& \text{Cathy}& \text{Charles}\\ \hline
\text{Name chosen}&\text{Bob}&\text{Brenda}&\text{Charles}&\text{Cathy}&\text{Adam}&\text{Alice}\\
\end{array}
$$
There are a total of $6!=720$ "named" permutations and exactly 80 of them
correspond to the situation where nobody gets their own name or the name of their partner.
You see that every permutation of "first letters" corresponds to exactly $2!\times 2!\times 2!=8$
permutations of "names". That's because you could write, in either order, "Adam, Alice" or "Alice, Adam" for the $A$s. That is, you have two ways to write in the A-names. Similarly there are two ways to write in the B-names and two ways to write in the C-names. There are eight times as many named permutations as un-named permutations.
So what is the right answer: 10 or 80? Well, that's up to you. An argument could be made for either choice, but once you know one of them, it is easy to calculate the other one.
Best Answer
$(i)$ Your answer is correct.
$(ii)$ This involves a derangement: an arrangement with no object in its proper place.
Denoting a derangement of $n$ objects as $D_n$, applying inclusion-exclusion,
and without bothering to further simplify the formula,
$D_6 = 6! -$ at least one gets right bag + at least two get right bag - ....
$= 6! - \dbinom61\cdot5! + \dbinom62\cdot4! -\dbinom63\cdot3! + ...... + \dbinom66\cdot0!$
$(iii)$
This is what is called a partial derangement, here simply $\binom61\cdot D_5$
PS
A simple formula for $D_n$ is to compute $\dfrac{n!}{e}$ and round to nearest integer.