Solution:

Let $a, b, c, d, e, f$ be the sides of the hexagon with $a=5$, $b=3$, $c=6$, $d=7$. Extend the sides $a, c, e$ until they meet at points $B, C, A$ (see figure). Since the interior angles of the hexagon are all $120^{\circ}$, the triangle $ABC$ and the three small triangles determined by each of the sides $f, b, d$ and respectively by the vertices $A, B, C$ are all equilateral, since all their angles are $60^{\circ}$. We therefore have that
$$b + c + d = d + e + f = f + a + b$$
from which, substituting the known values, we easily find that $f=8$ and $e=1$.

Solution:

Denoting the sides as in the previous solution, let us draw the perpendiculars to side $a$ through points $E$ and $J$; let $A, D$ and $I, F$ be the intersections of these perpendiculars with the lines of sides $a$ and $d$ respectively (see figure). Since the interior angles of the hexagon are all $120^{\circ}$, all the triangles in the figure have angles of $60^{\circ}, 30^{\circ}$ and $90^{\circ}$; hence $AI$ and $DF$ are also perpendicular to $IF$; it follows that $AD$ is parallel to $IF$, and the quadrilateral $ADFI$ is a rectangle as shown in the figure.

With straightforward calculations we find that
$$\begin{gathered} AD = AB + BC + CD = \frac{f}{2} + a + \frac{b}{2}, \quad IF = IH + HG + GF = \frac{e}{2} + d + \frac{c}{2}, \\ AI = AJ + JI = \frac{\sqrt{3}}{2}(f+e), \quad FD = FE + ED = \frac{\sqrt{3}}{2}(c+b) . \end{gathered}$$
Substituting the known values of $a, b, c, d$, and imposing the equalities $AD = IF$, $AI = FD$, we obtain $f-e=7$, $f+e=9$, hence $f=8$ and $e=1$.