3. Let $x_{k}=\tan \frac{k \pi}{17}(k=1,2, \cdots, 16)$. Then $\sum_{k=1}^{16} \frac{1}{1+x_{k}^{2}}=$ $\qquad$
3. $\frac{15}{2}$.

Notice,
$$
\begin{array}{l}
1+x_{k}^{2}=1+\tan ^{2} \frac{k \pi}{17}=\frac{1}{\cos ^{2} \frac{k \pi}{17}} . \\
\text { Then } \sum_{k=1}^{16} \frac{1}{1+x_{k}^{2}}=\sum_{k=1}^{16} \cos ^{2} \frac{k \pi}{17} \\
=\sum_{k=1}^{16} \frac{1+\cos \frac{2 k \pi}{17}}{2} \\
=8+\frac{1}{2} \sum_{k=1}^{16} \cos \frac{2 k \pi}{17} . \\
\text { By } 2 \sin \frac{\pi}{17} \cdot \sum_{k=1}^{16} \cos \frac{2 k \pi}{17} \\
=\sum_{k=1}^{16}\left(\sin \frac{(2 k+1) \pi}{17}-\sin \frac{(2 k-1) \pi}{17}\right) \\
=\sin \frac{33 \pi}{17}-\sin \frac{\pi}{17}=-2 \sin \frac{\pi}{17} \\
\Rightarrow \sum_{k=1}^{16} \cos \frac{2 k \pi}{17}=-1 . \\
\text { Therefore, the original expression }=8+\frac{1}{2}(-1)=\frac{15}{2} .
\end{array}
$$