So take $\sigma = (4,5)(2,3,7)$. The order, by definition, is the the smallest natural number $n$ such that $\sigma^n = (1)$ (i.e. the identity element in the group, i.e. the element that sends every number to itself). Since the cycles $(4,5)$ and $(2,3,7)$ are disjoint you have
$$\begin{align}
\sigma^n &= (4,5)(2,3,7)(4,5)(2,3,7)\dots (4,5)(2,3,7)(4,5)(2,3,7)\\
&= (4,5)(4,5)\dots (4,5)(4,5)(2,3,7)(2,3,7)\dots (2,3,7)(2,3,7)\\
&= (4,5)^n(2,3,7)^n
\end{align}$$
(note that the two elements $(4,5)$ and $(2,3,7)$ commute).
So the order of $\sigma$ is exactly the smallest natural number $n$ such that $(4,5)^n = (1)$ and $(2,3,7)^n = (1)$ (think about this fact for a moment).
But what is the order of a each of $(4,5)$ and $(2,3,7)$?
Well, the order of $(4,5)$ is two exactly because $(4,5)^2 = (4,5)(4,5) = (1)$. The order of $(2,3,7)$ is $3$ because
$$\begin{align}
(2,3,7)^1 &= (2,3,7) \\
(2,3,7)^2 &= (2,7,3) \\
(2,3,7)^3 &= (1)
\end{align}
$$
Now it is not to hard to see that the order of $\sigma$ is exactly the least common multiple of $2$ and $3$ (since we need both $(4,5)^m = (1)$ and $(2,3,7)^m = (1)$ and the smallest $m$ where this happens is exactly the least common multiple). Hence the final answer is $6$.
Addendum: I just wanted to add a bit about orders of these elements. First note that for example the element $(4,5)$ is just the element
$$
(4,5) = \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix}.
$$
(Hence $4$ maps to $5$ and $5$ to $4$). So when you compose (multiply) the element with itself, then you get
$$
\begin{align}
(4,5)(4,5) &= \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix}\begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8
\end{pmatrix} \\ &= \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &4 & 5& 6 & 7 & 8
\end{pmatrix} = (1)
\end{align}
$$
(I usually write the identity as $(1)$).
This means that the order of $(4,5)$ is $2$. Likewise you find that
$$
(2,3,7) = \begin{pmatrix}
1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 3 & 7 &4 & 5& 6 & 2 & 8
\end{pmatrix}.
$$
You are definitely on the right track. As you know that $lcm(3,4,5)=60$, you know that the following permutation has order $60$:
(1 2 3)(4 5 6 7)(8 9 10 11 12)
So $S_{12}$ definitely has an element of order $60$.
However, $3$, $4$, $5$ isn't the only set of numbers with an lcm of $60$. For example, $5$, $12$ also works. But that would give an element of $S_{17}$, which isn't any better.
You just need to do a little bit of explaining of various cases to show that any other set of numbers with an LCM of $60$ can't have a sum of less than $12$.
Best Answer
If $\sigma\tau = \tau\sigma$, then $$(\sigma\tau)^k = \sigma^k \tau^k$$ for any $k \in \mathbb N$. Therefore we can say that the order of $\phi = \sigma\tau$ divides the lowest common multiple of the orders of $\sigma$ and $\tau$.
However, it is false in general the the order of $\phi$ will equal this lowest common multiple.
For example, if $\sigma \in \langle \tau \rangle$, then $\tau$ and $\sigma$ will always commute, but the order of $\sigma\tau$ could be less than the lowest common multiple. In $S_3$, $(123)$ and $(132)$ both commute and have order $3$, but their product is $e$ which has order $1$.