3. D. Make $M P \perp A B$ at $P$, connect $P N$, then $P N \perp A B$. Therefore, $\angle M P N$ is the plane angle of the dihedral angle $A B C D-A B-A B E F$, $\angle M P N=120^{\circ}$.

Let $A M=F N=x$, then $M P=\frac{\sqrt{2}}{2} x, P N=\frac{\sqrt{2} a-x}{\sqrt{2}}, M N^{2}=M P^{2}+P N^{2}-2 M P \cdot P N \cdot \cos 120^{\circ}=$ $\frac{1}{2}\left(x^{2}-\sqrt{2} a x+2 a^{2}\right)$.
According to the problem, $0 \leqslant x \leqslant \sqrt{2} a$, hence when $x=\frac{\sqrt{2}}{2} a$, $(M N)_{\text {min }}=\frac{\sqrt{3}}{2} a$;
when $x=0$ or $\sqrt{2} a$, $(M N)_{\max }=a$.
3. For squares $A B C D$ and $A B E F$ with side length $a$, the planes they lie in form a $120^{\circ}$ angle. Points $M$ and $N$ are on the diagonals $A C$ and $B F$ respectively, and $A M = F N$. The range of values for $M N$ is ( ).
A. $\left(\frac{\sqrt{2}}{2} a, a\right]$
B. $\left(\frac{\sqrt{3}}{2} a, a\right]$
C. $\left[\frac{\sqrt{2}}{2} a, a\right]$
D. $\left[\frac{\sqrt{3} a}{2}, a\right]$