3. As shown in Figure 1,

A tangent line is drawn from a point $M$ outside the circle $\odot O$, touching $\odot O$ at point $B$. Line $M O$ intersects $\odot O$ at point $A$. Given that $M A=4$, $M B=4 \sqrt{3}$, and $N$ is the midpoint of arc $\overparen{A B}$. Then the area of the curvilinear triangle (shaded part) is $\qquad$
$3.8-\frac{4 \pi}{3}$.
According to the conditions, extend $M O$ to intersect $\odot O$ at point $C$. Let the radius of $\odot O$ be $r$. Then $M C=4+2 r$.
By the secant-tangent theorem, we have
$$
M B^{2}=M A \cdot M C,
$$

which is $48=4(4+2 r)$.
Solving for $r$ gives $r=4$.
Therefore, $O C=O A=A M=4$.
Connect $O B$. In the right triangle $\triangle O B M$,
$$
\begin{array}{l}
\sin \angle M O B=\frac{M B}{O M}=\frac{4 \sqrt{3}}{8}=\frac{\sqrt{3}}{2} \\
\Rightarrow \angle M O B=60^{\circ} .
\end{array}
$$

Thus, the degree measure of arc $\overparen{A B}$ is $60^{\circ}$, and since $N$ is the midpoint of arc $\overparen{A B}$, the degree measure of arc $\overparen{A N}$ is $30^{\circ}$.
Connect $O N$. Then $\angle M O N=30^{\circ}$. Therefore,
$$
\begin{array}{l}
S_{\triangle M O N}=\frac{1}{2} O M \cdot O N \sin 30^{\circ} \\
=\frac{1}{2} \times 8 \times 4 \times \frac{1}{2}=8 .
\end{array}
$$

The area of sector $A O N$ is
$$
\pi \times 4^{2} \times \frac{30}{360}=\frac{4 \pi}{3} \text {. }
$$

Therefore, the area of the shaded region is $8-\frac{4 \pi}{3}$.