Suppose that you’ve picked a legitimate set $A=\{a_1,a_2,a_3,a_4\}$, where $a_1<a_2<a_3<a_4$. Let $b_1=a_2-a_1$, the size of the gap between $a_1$ and $a_2$. Similarly, let $b_2=a_3-a_2$ and $b_3=a_4-a_3$. If you know the numbers $b_1,b_2$, and $b_3$, you almost know $A$. You would know $A$ if you also had $a_1$; then you could recover the whole lot:
$$\begin{align*}
a_1&=a_1\\
a_2&=a_1+b_1\\
a_3&=a_1+b_1+b_2\\
a_4&=a_1+b_1+b_2+b_3
\end{align*}$$
Now we know how to count the integer solutions to equations of the form $x_1+\ldots+x_m=n$, even subject to various conditions on the $x_k$, so it would be nice to be able to recast this problem in such terms. We’re not there yet, because $a_1+b_1+b_2+b_3$ doesn’t have to be a particular number. However, we could use another number, $b_4$, to measure the size of the gap from $a_4$ to $10$, the maximum possible value, by setting $b_4=10-a_4$. Now we know that $a_1+b_1+b_2+b_3+b_4=10$, and all four members of $A$ are recoverable from these five numbers.
Your instructor made one further small modification: instead of using $a_1$ directly, he used $a_1-1$, calling it $b_0$. You can think of $b_0$ as the gap between $1$, the lowest possible value, and the actual value of $a_1$. Notice that $b_0+b_1+b_2+b_3+b_4=(a_1-1)+b_1+b_2+b_3+b_4=10-1=9$, and that we can recover $A$ just as well from $b_0,b_1,b_2,b_3,b_4$ as from $a_1,b_1,b_2,b_3,b_4$:
$$\begin{align*}
a_1&=1+b_0\\
a_2&=1+b_0+b_1\\
a_3&=1+b_0+b_1+b_2\\
a_4&=1+b_0+b_1+b_2+b_3
\end{align*}$$
(We don’t need $b_4$ to reconstruct $A$; it’s there to make the sum a known constant.)
Thus, there is a bijection between $4$-element subsets $A$ of $\{1,\dots,10\}$ satisfying the condition that no two members are consecutive, and ordered $5$-tuples $\langle b_0,b_1,b_2,b_3,b_4\rangle$ of integers satisfying the equation $$b_0+b_1+b_2+b_3+b_4=9\tag{1}$$ together with the conditions $b_1,b_2,b_3\ge 2$ (to ensure that no two members of $A$ are consecutive), $b_0\ge 0$ (to ensure that $1$ is the smallest possible value of $a_1$), and $b_4\ge 0$ (to ensure that $10$ is the largest possible value of $a_4$.
Now the fact is that I wouldn’t use generating functions to count the number of solutions to $(1)$ subject to the given constraints. This is a bog-standard stars-and-bars problem that can be solved immediately by the technique explained in the article to which I linked. Because I think that the idea is a useful one to understand, I’ll give my own explanation here. Imagine that I have $9$ indistinguishable marbles and five boxes, labelled $B_0$ through $B_4$, and I want to count the number of distinguishable ways to put the $9$ marbles into the $5$ boxes. If you think of $b_k$ as representing the number of marbles in box $B_k$, you can see that the answer is the same as the number of solutions of $(1)$ in non-negative integers. Ah, but I’m required to have at least $2$ marbles in each of boxes $B_1,B_2$, and $B_3$. No problem: I just put $2$ marbles into each of those boxes right away, leaving myself with just $9-6=3$ marbles to distribute as I please amongst the $5$ boxes. (By the way, this shows that the number of solutions of $(1)$ that meet the stated conditions is the same as the number of solutions of $$b_0+b_1+b_2+b_3+b_4=3\tag{2}$$ in non-negative integers.) From here on I’ll simply ignore the $6$ required marbles and look at the number of ways of distributing the other three.
Draw a schematic of the boxes: instead of drawing the boxes themselves, draw the boundaries between them, representing them by vertical bars, $|$: $$|\;|\;|\;|\;.$$ The spaces between the bars, before the first bar, and after the last bar represent the five boxes. If I represent the marbles by $o$’s, I can symbolize the arrangement with two marbles in the first box and one in the third by $$oo||o||\;.$$ Each possible distribution of the three marbles corresponds to a unique string of three $o$’s and four $|$’s, and each string of three $o$’s and four $|$’s corresponds to a unique distribution of the three marbles. Thus, the number of solutions to $(2)$ in non-negative integers, and hence the number of solutions to $(1)$ that meet the extra conditions, is simply the number of strings of four $|$’s and three $o$’s. This is just the number of ways of choosing which $3$ of the $7$ positions in the string are to be $o$’s, so it’s $\binom73$. (Or, if you prefer, it’s the number of ways of choosing which $4$ of the $7$ position in the string are to be $|$’s, which is $\binom74$; of course these are the same.)
If you use generating functions directly on $(1)$, it works like this. The possible values for $b_0$ and $b_4$ are $0,1,\dots$, while those for $b_1,b_2$, and $b_3$ are $2,3,\dots$. These are the exponents on the terms of $\sum\limits_{n\ge 0}x^n$ and $\sum\limits_{n\ge 2}x^n$, respectively. We know that formally $$\sum_{n\ge 0}x^n=\frac1{1-x}\;,$$ so $$\sum_{n\ge 2}x^n=x^2\sum_{n\ge 0}x^n=\frac{x^2}{1-x}\;.$$ Let $$g(x)=\left(\sum_{n\ge 0}x^n\right)^2\left(\sum_{n\ge 2}x^n\right)^3=\frac{x^6}{(1-x)^5}\;;$$ this is the desired generating function.
To see what this means, imagine multiplying out the summations:
$$(1+x+x^2+x^3+\ldots)^2(x^2+x^3+x^4+\ldots)^3=x^6(1+x+x^2+x^3+\ldots)^5\;.$$
When you’ve collected terms, the coefficient of $x^9$ will be the number of products that produced $x^9$: $x^6\cdot x^0\cdot x^0\cdot x^0\cdot x^0\cdot x^3$, $x^6\cdot x\cdot x^0\cdot x^0\cdot x^2\cdot x^0$, etc. Each of these corresponds to one solution of $(1)$. For instance, the two that I just listed correspond to $$0+2+2+2+3=9$$ and $$1+2+2+4+0=9\;,$$ respectively. Thus, the answer to the question is simply the coefficient of $x^9$ in $$g(x)=\frac{x^6}{(1-x)^5}\;.$$ This is the coefficient of $x^3$ in $$\frac1{(1-x)^5}=(1+x+x^2+x^3+\ldots)^5\;,$$ which is the same as the coefficient of $x^3$ in $$(1+x+x^2+x^3)^5\;,$$ since the higher powers of $x$ can’t contribute to an $x^3$ term. You can go ahead and work out this coefficient by direct calculation, though it’s a bit tedious.
First we add subsets by taking any of the $9$ pairs of consecutive integers $\{1,2\},\{2,3\},...,\{9,10\}$ and an arbitrary choice of the third element - in each case there are $8$ such choices, so this gives $9 \cdot 8 = 72$.
However, we see that any set of three consecutive integers $\{1,2,3\},\{2,3,4\},...,\{8,9,10\}$ has been counted twice, so we remove these. There are $8$ of them, so we subtract that and get $9\cdot 8 - 8 = 64$.
Best Answer
For i.): Choosing a certain number $m$ of elements out of a set of $n$ without regard for the order is calculated by $n \choose m$. In this case it is $12 \choose 6 $$=924$.
For ii.): Possibilities to get 4 even numbers out of 6 is $6 \choose 4$$=15$ and there are $6 \choose 2$$=15$ ways to get 2 odd numbers out of 6 (which is by the way the same number because you can reformulate it to not count the ways to draw 4 but to throw out 2). To get the final number of ways we have to multiplicate both, so we get $15\cdot 15=225$.
For iii.): There are 6 odd integers in the set, you decide for each odd number if you include it in a set or not, that is 2 choices - and as there are 6 odd numbers you get to choose 6 times. You can get each of the described sets with this method and you miss none if you play through all choices. So the total number of sets containing only those is $2^6=64$.