Your answer is correct if you were trying to write what I think you were. You were doing two row notation and multiplying two permutations?

To do the second they are written in cycle notation. This can be confusing at first because of the order. We compose functions backwards (i.e. you apply g first then f), and the same goes with permutations. So within the parenthesis you read right to left, but go from right to left in terms of the compositions.

The way you work through these is as follows.

(1 3 6)(1 2 6 5)(4 5)

Take 1 and see where it goes.

It is untouched by (4 5). (1 2 6 5) takes 1 to 2. then (1 3 6) does not affect 2.

so 1 -> 2.

Now you see where 2 goes.

It is untouched by (4 5). (1 2 6 5) takes 2 -> 6. Now (1 3 6) takes 6 -> 1. Notice the individual cycles "wrap around" so their representation is unique up to cycling through although generally you write the smallest number first.

so we see 2 -> 1.

This gives us the cycle (1 2).

Now proceed to 3. I will omit details

3 -> 6

6 -> 5

5 -> 4

4-> 5 -> 1 ->3

so we have (3 6 5 4)

Thus this multiplication is simply (1 2) (3 6 5 4).

Also remember if in general it is possible for a number i to be fixed. One can write (i) to represent it gets sent to itself, or more generally it is simply omitted from the cycle decomposition.