For question 2, there was a paper published in 2013, Standard Young tableaux and colored Motzkin paths.
http://www.sciencedirect.com/science/article/pii/S009731651300099X.
They used "colored Motzkin paths" to characterize standard Young tableaux, and I think "colored Motzkin paths" is something like non-crossing partition. Roughly, the level step is a fixed-point in the involution and the pairs of up and down steps are the cycle in the involution. The most important things in this paper is that they found the relation between "bounded height SYT" and some restrictions of "colored Motzkin path".
It might help to consider an example. Suppose we have the following disjoint cycle decomposition of a permutation:
$$
\sigma = (1 \ 7\ 4\ 10)(2\ 9\ 8)(3\ 5\ 6).
$$
To begin, separately decompose each cycle:
$$
(1 \ 7\ 4\ 10) = \tau_{1,1}\tau_{1,2}, \quad (2\ 9\ 8) = \tau_{2,1}\tau_{2,2} \quad (3\ 5\ 6) = \tau_{3,1} \tau_{3,2}.
$$
Note that $\tau_{1,1},\tau_{1,2}$ are permutations that only affect the elements $1,4,7,10$. Similarly, $\tau_{2,1},\tau_{2,2}$ only affect $2,9,8$. In other words, for any $i \neq j$, the elements $\tau_{i,p}, \tau_{j,q}$ are disjoint permutations, which means that $\tau_{ip}\tau_{jq} = \tau_{jq}\tau_{ip}$.
With that established, we can use this commutativity property to "move" the transpositions $\tau_{i,1}$ to the left. That is, we can write
$$
\sigma = \tau_{11}(\tau_{12}\color{red}{\tau_{21}})\tau_{22}\tau_{31}\tau_{32}\\
= \tau_{11} (\color{red}{\tau_{21}}\tau_{12}) \tau_{22}\color{red}{\tau_{31}} \tau_{32}\\
= \tau_{11}\tau_{21}\color{red}{\tau_{31}}\tau_{12}\tau_{22}\tau_{32}\\
= (\tau_{11}\tau_{21}\tau_{31})(\tau_{12}\tau_{22}\tau_{32}).
$$
Now, we see that $\sigma$ is a product of the involutions $\tau_{11}\tau_{21}\tau_{31}$ and $\tau_{12}\tau_{22}\tau_{32}$.
Best Answer
For your three numbered statements, $(1)$ is true (easy proof), but $(2)$ and $(3)$ are false. A counter example for $(2)$ is obtained by taking $X = (1,2)$ and $Y = (3,4)$. Then $XY = (1,2)(3,4)$ is an involution as well. A counterexample for $(3)$ is obtained from this example as well; the involution $X = (1,2)(3,4)$ can be factored as $YZ$ where $Y = (1,2)$ and $Z = (3,4)$.
As for the statement in question, here's a quick proof sketch:
(1) By using the disjoint cycle decomposition, you can reduce to proving that the cycle $(1,2,3,\dots,n)$ can be written as a product of two involutions in $S_n$.
(2) To handle that case, draw $n$ vertices in the plane (labelled $1,2,\dots,n$) and connect the $n$ vertices by drawing $n-1$ edges. This will make a unique (up to choice of direction to travel) path in your graph. Label the edges $1,2,\dots,n-1$ in the order of the path. For each edge, put the two vertices connected by that edge into a two-cycle. Then form $\pi_1$, the product of the two-cycles formed in this way from odd-numbered edges, and $\pi_2$, the product of the two-cycles formed in this way from even-numbered edges. Then the product $\pi_2 \pi_1$ is an $n$-cycle $\tau$. This needs to be checked; in fact, if you number the vertices in the order of the path, then $\tau = (1,3,5,\dots, 6,4,2)$. Conjugate the relation $\tau = \pi_2 \pi_1$ to get that $(1,2,\dots,n)$ is a product of two involutions.