5. As shown in Figure 1, in rhombus $A B C D$, it is known that $A B=3, D F$ $=1, \angle D A B=$ $60^{\circ}, \angle E F G=$ $15^{\circ}, F G \perp B C$. Then $A E=(\quad$.
(A) $1+\sqrt{2}$
(B) $\sqrt{6}$
(C) $2 \sqrt{3}-1$
(D) $1+\sqrt{3}$
5. D.

Draw a perpendicular from point $F$ to $AB$, with the foot of the perpendicular being $H$.
Given $\angle DAB = 60^{\circ}$ and $AF = AD - FD = 2$, we have
$$
\begin{array}{l}
\angle AFH = 30^{\circ}, AH = 1, FH = \sqrt{3}. \\
\text{Also, } \angle EFH = \angle AFG - \angle AFH - \angle EFG \\
= 45^{\circ}.
\end{array}
$$

Thus, $\triangle FHE$ is an isosceles right triangle.
Therefore, $HE = FH = \sqrt{3}$.
Hence, $AE = AH + HE = 1 + \sqrt{3}$.