$$\boxed{\frac{15}{8}}$$
Let $P$ be the point of intersection of lines $AD$, $BE$ and $CF$. Then, we have $\angle PBA = \angle PDE$, since they are subtended by the same arc $\overarc{EA}$ of the circle at the points $B$ and $D$ on the circumference. We also have $\angle BPA = \angle DPE$ so that the triangles $BPA$ and $DPE$ are similar. Consequently, we get
$$(1) \qquad PA : PE = BA : DE = 1 : 4.$$
In the same way, we see that the triangles $CPB$ and $EPF$ are similar, and therefore, we get
$$(2) \qquad PE : PC = EF : CB = 5 : 2.$$
From (1) and (2) above, we get $PC : PA = 8 : 5$. Also, from the similarity of the triangles $FPA$ and $DPC$, we obtain $PC : PA = DC : FA$, from which it follows that $FA = \frac{5}{8} \cdot DC = \frac{15}{8}$, which is the desired answer.