Finding the cosets in this small case is not so bad. First and foremost you have the coset
$$H=\{e,(123),(132)\}$$
Here is a general algorithm for how to finish. So far, say you've computed $k<8$ cosets. Pick an element of $S_4$ that is not in any of the cosets you've computed so far, and that will give you a new coset.
With this method I'll find one more coset for you. The cosets we have are
$$\{e,(123),(132)\}$$
$$\{(1234),(1234)(123)=(2413),(1234)(132)=(14)\}$$
None of the two cosets we've found contain $(234)$. So we multiply the elements of $H$ on the left by $(234)$ to get a new left coset,
$$\{(234),(24)(13),(142)\}$$
Every element of the group is in exactly one coset, so continuing in this way will get you to the end.
The permutation $(1)$ simply means the identity, which sends:
$1 \mapsto 1\\2 \mapsto 2\\3 \mapsto 3.$
We could also denote it by $(1)(2)(3)$ (three disjoint $1$-cycles), which might be a bit less confusing. It's common practice to omit $1$-cycles from permutations, but in the case of the identity map, that leaves us "nothing to write down" (which of course is what the identity map does-namely, "nothing").
If you apply the permutation $(1\ 2)$ to each element of $H$, which is what is meant by:
$(1\ 2)H$, you obtain (applying the pemutation on the right first, as is often done with composition):
$(1\ 2)(1) = (1\ 2)$
$(1\ 2)(1\ 2\ 3) = (2\ 3)$
$(1\ 2)(1\ 3\ 2) = (1\ 3)$, which is what you surmised.
It's not that hard to see that $\{(1), (1\ 2\ 3), (1\ 3\ 2)\}$ form a cyclic subgroup of order $3$, which we might also denote as:
$H = \{1,a,a^2\}$ (since $(1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2)$).
then $(1)H = 1H = H$ (this is obvious), while:
$(1\ 2\ 3)H = aH = \{a,a^2,a^3 = 1\}$, which is the same SET (the order of elements doesn't matter in a set) as $H$.
Similarly, $(1\ 3\ 2)H = a^2H = \{a^2,a^3 = 1,a^4 = a^3a = a\} = H$.
It turns out in this case (and you might well speculate as to what happens in general) that for $\sigma \in S_3,$ we have:
$\sigma H = H \iff \sigma \in H$
Best Answer
The left cosets of a subgroup $K$ of a group $G$ are defined to be
$$aK=\{ ak\mid k\in K\}$$
for each $a\in G$.
So, to get your coset, simply multiply each element of $H$ on the left by the chosen representative.
For example,
$$\begin{align} (12)H&=\{(12)(1), (12)(123), (12)(132)\}\\ &=\{(12), (1)(23), (13)(2)\}\\ &=\{(12), (13), (23)\}. \end{align}$$