$\iota$ represents the identity permutation: every element in $\{1, 2, ..., 12\}$ is mapped to itself:
$\quad \iota = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)$.
Recall the definition of the order of any element of a finite group.
In the case of a permutation $\pi$ in $S_n$, the exponent $k$ in your question represents the order of $\pi$.
That is, if $\;\pi^k = \underbrace{\pi\circ \pi \circ \cdots \circ \pi}_{\large k \; times}\; = \iota,\;$ and if $k$ is the least such positive integer such that $\pi^k = \iota$, then $k$ is the order of $\pi$.
A permutation expressed as the product of disjoint cycles has order $k$ equal to the least common multiple of the lengths of the cycles.
So, if $\pi = (1,2)(3,4,5,6,7)(8,9,10,11)(12) \in S_{12}$,
Then the lengths $n_i$ of the 4 disjoint cycles of $\pi$ are, in order of their listing above, $n_1 = 2, \; n_2 = 5, \; n_3 = 4,\;n_4 = 1.\;$.
So the order $k$ of $\pi$ is given by the least common multiple $$\;\text{lcm}\,(n_1, n_2, n_3, n_4) = \operatorname{lcm}(2, 5, 4, 1) = 20.$$ That is, $\pi^k = \pi^{20} = \iota,\;$ and there is NO positive integer $n<k = 20\,$ such that $\pi^n = \iota$.
What this means is that $$\underbrace{\pi \circ \pi \circ \cdots \circ \pi}_{\large 20 \; factors} = \iota$$ and $$\underbrace{\pi \circ \pi \circ \cdots \circ \pi}_{ n \; factors,\;1 \,\lt n\, \lt 20} \neq \iota$$
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
To answer the first question: (1,2,3,4)(5,6,7,8,9) for n=9 appears to be the smallest n.
A counter example to the second statement is provided by the even permutation (1,2,3)(4,5,6,7,8,9,10) for n=10.