Solution:

From $\angle E = 60^{\circ}$, we get that $\angle AEF = 120^{\circ} - \angle CED = \angle CDE$. Therefore, $\triangle AEF \sim \triangle CDE$. Since $EF:DE = 2:1$, the ratio of similarity must be $2:1$, so $AE = 2CD = 8$. Recall $ABC$ has side length $7 + 4 = 11$, so $EC = 11 - 8 = 3$. Law of Cosines on $\triangle CDE$ gives $DE^2 = \sqrt{3^2 + 4^2 - 3\cdot 4} = \sqrt{13}$.

Solution:

Let $\odot (DEF)$ meet $AC$ again at point $X$. Then, $\angle FXA = 180^{\circ} - \angle FXE = \angle FDE = 90^{\circ}$ and $\angle XDC = 180^{\circ} - \angle DCX - \angle DXC = 120^{\circ} - \angle DXE = 120^{\circ} - \angle DFE = 90^{\circ}$. It follows that $CX = 2CD = 8$, so $AX = 11 - CX = 3$, and $AF = 2AX = 6$. Thus, Law of Cosines on $\triangle AEF$ gives $EF = \sqrt{8^2 + 6^2 - 8\cdot 6} = 2\sqrt{13}$, implying that $DE = \sqrt{13}$.