Note that in cycle notation for permutations, $(1~3~4~5~7) = (5~7~1~3~4)$.
Part 1: The idea here would be to create a cycle that moved 1 to 3 after two moves, 3 to 4 after two moves, 4 to 5 after two moves, 5 to 7 after two moves, and 7 to one after two moves.
Then (arbitrarily and without loss of generality), 1 is first in our cycle notation, followed by something, then 3, then something, then 4. So we have $(1~a_1~3~a_2~4)$ as our cycle. Now we try to figure out what goes in place of $a_1$ and $a_2$. After 1 iteration of this permutation, 4 maps to 1. On the second iteration, it should map to five, so that $x^2(4) = 5$. Note that I'm viewing $x$ as a function with $x^2 = x \circ x$. We now have $x = (1~5~3~a_2~4)$. Next, we need $x^2(5) = 7$. The way to do that is to set $a_2 = 7$. Thus we see, and can confirm, that $(1~5~3~7~4)^2 = (1~3~4~5~7)$. There's a pair left over in $S_7$; we haven't moved 2 or 6. Since $(2~6)^2$ is the identity, we can set $x = (1~5~3~7~4)(2~6)$ as well.
Hopefully this helps; the teacher in me won't let me do the whole problem (or the actual proof) for you. :)