3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$

then $x^{2}+y^{2}+z^{2}+w^{2}=$
3. 36 .

It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$

which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$

Therefore,
$$
\begin{array}{l}
1^{2}+3^{2}+x^{2}+y^{2}=2^{2}+4^{2} \\
\Rightarrow x^{2}+y^{2}=2^{2}-1^{2}+4^{2}-3^{2}=10 .
\end{array}
$$

Similarly, $z^{2}+w^{2}=6^{2}-5^{2}+8^{2}-7^{2}=26$.
Thus, $x^{2}+y^{2}+z^{2}+w^{2}=36$.