I'm not entirely sure what you mean by (1)(2)(3)(4), is this just the identity permutation?
Notation aside, your answer is not correct. You need to take all permutations of order 4 and multiply them individually to find the subgroups generated by them. This will mean that every such group will contain an element of order 4. The subgroup you give contains the identity, two elements of order 2 and an element of order 3...You cannot have an element of order 3 as the order of an element must divide the order of a group. Here your subgroup (viewed as a group) has order 4, but 3 does not divide 4.
Now, as it happens all permutations of order 4 in have the same form (can you tell me what they look like?). So, take one of these elements and write down , , (and you know that ). This is one of your cyclic subgroups, the others look identical, just with a different permutation to start with.
For the second part, pick an element, say `a'. What does the set of all permutation in which fix `a' look like?