One, (Full marks 20 points) Given positive numbers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}+c^{2}=4^{2}, \\
b^{2}+c^{2}=5^{2} .
\end{array}\right.
$$

Find the range of $a^{2}+b^{2}$.
Let $a^{2}+b^{2}=k$. From the given information, we have
$$
b^{2}-a^{2}=\left(b^{2}+c^{2}\right)-\left(a^{2}+c^{2}\right)=5^{2}-4^{2}=9 .
$$

Solving this, we get $a^{2}=\frac{k}{2}-9$
$$
\begin{array}{l}
i^{2}=\frac{k+9}{2}>9, \\
c^{2}=\frac{41-k}{2}>0 .
\end{array}
$$

Thus, $9<k<41$.
That is, the value of $a^{2}+b^{2}$ will not exceed $(9,41)$.
Conversely, for $k$ satisfying (4), there exist positive numbers $a, b, c$ that satisfy (1), (2), and (3). From (1) + (3) and (2) + (3), we get
$$
\left\{\begin{array}{l}
a^{2}+c^{2}=4^{2} \\
b^{2}+c^{2}=5^{2}
\end{array}\right.
$$

which fits the given conditions.
Therefore, all numbers in $(9,41)$ are values of $a^{2}+b^{2}$.
In summary, the range of values for $a^{2}+b^{2}$ is $(9,41)$.
Analysis: This problem has a clear geometric background. As shown in Figure 6, on the three edges of a cube sharing a vertex, take $V A=a, V B=b, V C=c$, then
$$
\begin{array}{l}
A B=\sqrt{a^{2}+c^{2}}=4, \\
B C=\sqrt{b^{2}+c^{2}}=5, \\
C A=\sqrt{a^{2}+b^{2}} .
\end{array}
$$

By solid geometry knowledge, $\triangle A B C$ is an acute triangle. As shown in Figure 7, with $B A=4$ as the radius and point $B$ as the center, draw a $\frac{1}{4}$ circle such that $\angle D B C=90^{\circ}$, then $B E=B D=4$. Draw the tangent $C F$ and the secant $C D$ from $C$. Then
$$
C F=3, C D=\sqrt{41} \text {. }
$$

Since $\triangle A B C$ is an acute triangle, $A$ lies in
$$
C F<C A<C D \text {, }
$$

which means $3<C A<\sqrt{41}$,
leading to $9<a^{2}+b^{2}<41$.