3.13. From vertex $C$ of the rhombus $A B C D$, whose side is equal to $a$, two segments $C E$ and $C F$ are drawn, dividing the rhombus into three equal-area figures. It is known that $\cos C=\frac{1}{4}$. Find the sum $C E+C F$.
3.13. The heights of triangles $C E D$ and $C F B$, drawn from vertex $C$ (Fig. 3.15), have equal lengths and $S_{\triangle C E D} = S_{\triangle C F B}$ (by condition); therefore, $D E = F B$, and thus $C E = C F$ and $A E = A F$. We draw the diagonal $A C$, which divides the quadrilateral $A E C F$ into two equal triangles. Therefore, $S_{\triangle A C F} = \frac{1}{2} S_{A E C F}$. Since $S_{A E C F} = S_{\triangle C F B}$ (by condition), then $S_{\triangle A C F} = \frac{1}{2} S_{\triangle C F B}$, and the triangles $A F C$ and $F C B$ have a common height drawn from vertex $C$. It follows that $A F = \frac{1}{2} F B$, i.e., $F B = \frac{2}{3} a$; moreover, $\cos B = \cos \left(180^{\circ} - C\right) = -\cos C = -\frac{1}{4}$. From $\triangle F C B$ using the cosine theorem, we get

$$
C F = \sqrt{a^{2} + \frac{4 a^{2}}{9} - 2 a \cdot \left(-\frac{1}{4}\right) \cdot \frac{2 a}{3}} = \frac{4 a}{3}
$$

Thus, $C E + C F = \frac{8 a}{3}$.