8. As shown in Figure 3, in $\square A B C D$, $M$ and $N$ are the midpoints of sides $B C$ and $D C$ respectively, $A N=1, A M=2$, and $\angle M A N=$ $60^{\circ}$. Then the length of $A B$ is $\qquad$.
8. $\frac{2 \sqrt{13}}{3}$.

As shown in Figure 7, extend $A M$ to intersect the extension of $D C$ at point $F$. It is easy to prove that $\triangle A M B \cong \triangle F M C$, so $C F=A B$. Therefore, $N F=\frac{3}{2} A B$.

Draw a perpendicular from point $N$ to $A F$ at $H$, then $A H=\frac{1}{2} A N=\frac{1}{2}, N H=\frac{\sqrt{3}}{2}$.
Thus, $H F=2 \times 2-\frac{1}{2}=\frac{7}{2}$,
$$
N F=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{7}{2}\right)^{2}}=\sqrt{13} \text {. }
$$

Therefore, $A B=\frac{2}{3} N F=\frac{2 \sqrt{13}}{3}$.