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 $2^6=64$.