If $\sigma$ and $\tau$ are disjoint cycles, then $\sigma\tau=\tau\sigma$; in particular, for every integer $k$, we have $(\sigma\tau)^k = \sigma^k\tau^k$.
So the first observation is that you can compute $a^{14}$ by computing $(1\ 2\ 5)^{14}$, $(3\ 8\ 4)^{14}$, and $(6\ 7)^{14}$ separately.
The second observation is that if $\sigma^n$ is the identity, and $a\equiv \pmod{n}$, then $\sigma^a = \sigma^b$.
Since $(1\ 2\ 5)^{3}$, $(3\ 8\ 4)^3$, and $(6\ 7)^2$ are the identity, you only need to compute $(1\ 2\ 5)^k$ where $k$ is any number congruent to $14$ modulo $3$; $(3\ 8\ 4)^s$ where $s$ is any number congruent to $14$ modulo $3$; and $(6\ 7)^m$ where $m$ is any number congruent to $14$ modulo $2$.
I'll leave that to you.