Consider the symmetric and alternating groups $S_n$ and $A_n$ ($n>2$).
1. Does arbitrary $2$-cycle and an arbitrary $n$-cycle in $S_n$ generates $S_n$?
2. If $n$ is odd, does an arbitrary $3$-cycle and arbitrary $n$-cycle in $S_n$ generates the subgroup $A_n$?
3. What are the references which give various "presentations" of $S_n$, $A_n$, and about order of products in $S_n$?.
Besides to @Alexander Answer, I can find some points among my old notes. I hope they help you.
It is noted that every elemnt of $S_n$ is as product of distict cycles and so $S_n$ can be generated by $(i,i+1),~~i=1,2,...,n-1$. In fact for any $(i,j)\in S_n$: $$(i,j)=(i,i+1)(i+1,i+2)...(j-1,j)(j-2,j-1)...(i+1,i+2)(i,i+1)$$ In other words, $$S_n=\langle(i,i+1)\mid~i=1,2,...,n-1\rangle$$ Moreover:
$(i,i+1)^2=1,~i=1,2,...,n-1~~~~$ and $~~~~ \Big((i,i+1)(i+1,i+2)\Big)^3=1,~i=1,2,...,n-2$
and $~~(i,i+1)(j,j+1)=(j,j+1)(i,i+1),~~i,j=1,2,...,n-1,~|i-j|\ge2$
so if we set $$P_n=\langle x_1,x_2,...x_{n-1}\mid R,S,T\rangle$$ wherein $R=\{x_i^2\mid i=1,...,n-1\}$, $S=\{(x_i,x_{i+1})^3\mid i=1,...,n-1\}$ and $T=\{[x_i,x_j]\mid 1\leq i\leq j-1<n-1\}$ then by using the following bijection $S_n$ gets a presentation as above: $$\theta: P_n\to S_n\\ x_i\to(i,i+1)$$
There is a stared problem in Dixon's book as follows:
$^*2.63.$ Let $x$ be any nontrivial element of $S_n$. If $n\neq 4$, then there exists $y\in S_n$ such that $S_n=\langle x,y\rangle$.
Consider $(12)$ and $(1324)$ in $S_4$.
This is true. Hint: We know that $A_n$ is generated by the set of $3$-cycles in $S_n$. What happens when we conjugate a $3$-cycle by an $n$-cycle? Why is it important that $n$ is odd? As Derek Holt points out my argument contains a careless error. This statement is false too.
Here's one reference about presentations. Orders of products of permutations are not well understood in the general sense, as far as I know. (Though perhaps someone can provide improve on this point for you.)