I'll use a longer cycle to help describe two techniques for writing disjoint cycles as the product of transpositions:
Let's say $\tau = (1, 3, 4, 6, 7, 9) \in S_9$
Then, note the patterns:
Method 1: $\tau = (1, 3, 4, 6, 7, 9) = (1, 9)(1, 7)(1, 6)(1, 4)(1, 3)$
Method 2: $\tau = (1, 3, 4, 6, 7, 9) = (1, 3)(3, 4)(4, 6)(6, 7)(7, 9)$
Both products of transpositions, method $1$ or method $2$, represent the same permutation, $\tau$. Note that the order of the disjoint cycle $\tau$ is $6$, but in both expressions of $\tau$ as the product of transpositions, $\tau$ has $5$ (odd number of) transpositions. Hence $\tau$ is an odd permutation.
Now, don't forget to multiply the transpositions you obtain for each disjoint cycle so you obtain an expression of the permutation $S_{11}$ as the product of the product of transpositions, and determine whether it is odd or even:
$\sigma = (1, 4, 10)(3, 9, 8, 7, 11)(5, 6)$.
In product notation the permutations are normally applied from right to left. Be sure to distinguish the product of two permutations from the concatenation of cycles within a single permutation, by using a dot ($\cdot$) to signify 'product'.
Verify the second example on a string like $abcd$:
$$
abcd \stackrel{(243)}\to acdb\stackrel{(1243)}\to dabc
$$
is the same as:
$$
abcd\stackrel{(14)}\to dbca\stackrel{(34)}\to dbac\stackrel{(23)}\to dabc.
$$
This proves that $(1243)\cdot(243)=(23)\cdot(34)\cdot(14)$.
You have a typo in your first example. The assertion $(132)=(13)\cdot(12)$ is false, since
$$
abc\stackrel{(132)}\to bca
$$
while
$$
abc\stackrel{(12)}\to bac\stackrel{(13)}\to cab
$$
But it is true that $(132)=(12)\cdot(13)$:
$$
abc\stackrel{(13)}\to cba\stackrel{(12)}\to bca.$$
The product of a transposition with itself is the identity. The transposition $(ij)$ swaps element $i$ with element $j$. Doing this a second time will return the elements to their original places.
Best Answer
First, we have $a=(1)(23847)(56)$. The next thing to note is that if $\sigma=(i_1i_2\cdots i_n)$ is a cycle, we can express it as a product of transpositons as $$ \sigma=(i_1i_2)(i_2i_3)\cdots(i_{n-1}i_n). $$ In this example, you can verify that $$ (23847)=(23)(38)(84)(47). $$ Hence, $a=(23847)(56)=(23)(38)(84)(47)(56)$. It is a product of an odd number of transpositions, making it an odd permutation.