Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$, where $\sigma\in S_{14}$ and $|\sigma|=28$? 
Let $\sigma\in S_{14}$ which is an even permutation of the order of $28$. Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$.

My try:
We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and $28=2\cdot2\cdot7$.
So $\sigma$ must be character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ - composition of row cycle $4$ and row cycle $7$
 because if $\sigma$ would be a character $(a_1
,a_2)(b_1b_2)(c_1...c_7)$ then $|\sigma|=2\cdot7=14$ which is contrary to the assumption.
That's why $4+7=11$ elements elements undergo nontrivial permutations and $14-11=3$ elements pass on to each other.
So $\sigma$ has a character:
$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$
Moreover we have information that $\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ is composition an even number of transpositions.
However these are my only thoughts and I don't know what to do next to come to the thesis.
EDIT:
According to the remark from @EricTowers $\sigma$ can still have a character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$ then $$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$

You already have the prime factorisation of $28$. To get an element of order $28$, you need to partition $14$ into divisors of $28$ (namely, $1$, $2$, $4$, $7$, and $14$) so that their LCM is $28$.${}^\dagger$ So, what are the partitions of $14$ into those divisors, potentially including $1$, $4$, and $14$, such that the disjoint cycles of elements of $S_{14}$ form elements of order $28$ with cyclic decompositions composed of those divisors?
You'll find that you'll always need a $1$ in the cyclic decomposition. What does that imply?
You need to have at least one term of $7$ or $14$ in the partition. It should be obvious why you can't have a term $14$; can you have two terms of $7$? If the number of $7$s in the partition is odd, what does that say about the number of $1$s in the partition?${}^\dagger$
$\dagger$: I am thankful to @StevenStadnicki for the clarifying sentences provided in the comments.
