3. As shown in Figure 4, in quadrilateral $ABCD$, let $\angle BAD + \angle ADC = 270^{\circ}$, and $E, F$ are the midpoints of $AD, BC$ respectively, $EF=4$, the shaded parts are semicircles with diameters $AB, CD$ respectively. Then the sum of the areas of these two semicircles is $\qquad$ (the value of pi is $\pi$).
$3.8 \pi$. 
As shown in Figure 7, extend $B A$ and $C D$ to intersect at point $M$.

From $\angle B A D + \angle A D C = 270^{\circ}$, we get
$$
\begin{array}{l}
\angle B M C \\
= \angle A M D = 90^{\circ}.
\end{array}
$$

Connect $B D$ and take the midpoint $P$ of $B D$, then connect $P E$ and $P F$.
By the midline theorem of a triangle, we have
$$
P E \parallel \frac{A B}{2}, P F \parallel \frac{C D}{2}.
$$

Therefore, $\angle E P F = \angle B M C = 90^{\circ}$.
Applying the Pythagorean theorem in the right triangle $\triangle E P F$, we get
$$
P E^{2} + P F^{2} = E F^{2} = 4^{2},
$$

which means $\left(\frac{A B}{2}\right)^{2} + \left(\frac{C D}{2}\right)^{2} = 16$.
Thus, the sum of the areas of the two shaded semicircles is
$$
\frac{\pi}{2} \left(\frac{A B}{2}\right)^{2} + \frac{\pi}{2} \left(\frac{C D}{2}\right)^{2} = \frac{\pi}{2} \times 16 = 8 \pi.
$$