14. (1) As shown in Figure 9 (a), within the square $ABCD$, two moving circles $\odot O_{1}$ and $\odot O_{2}$ are externally tangent to each other, and $\odot O_{1}$ is tangent to sides $AB$ and $AD$, while $\odot O_{2}$ is tangent to sides $BC$ and $CD$. If the side length of the square $ABCD$ is $1$, and the radii of $\odot O_{1}$ and $\odot O_{2}$ are $r_{1}$ and $r_{2}$ respectively, find:
(i) the relationship between $r_{1}$ and $r_{2}$;
(ii) the minimum value of the sum of the areas of $\odot O_{1}$ and $\odot O_{2}$.
(2) As shown in Figure 9 (b), if the square $ABCD$ in (1) is changed to a rectangle with a width of 1 and a length of $\frac{3}{2}$, and all other conditions remain the same, does the sum of the areas of $\odot O_{1}$ and $\odot O_{2}$ have a minimum value? If not, explain why; if so, find this minimum value.
14. (1) (i) As shown in Figure 15, in the square $ABCD$, connect $AC$. Clearly, points $O_1$ and $O_2$ lie on $AC$, and
$$
\begin{array}{l}
AO_1 = \sqrt{2} r_1, \\
O_1O_2 = r_1 + r_2, \\
CO_2 = \sqrt{2} r_2
\end{array}
$$

From $AC = AO_1 + O_1O_2 + CO_2 = \sqrt{2}$, we get
$$
\sqrt{2} r_1 + r_1 + r_2 + \sqrt{2} r_2 = \sqrt{2}.
$$

Thus, $r_1 + r_2 = 2 - \sqrt{2}$.
(ii) It is easy to see that $r_1 \leq \frac{1}{2}$ and $r_2 \leq \frac{1}{2}$. Therefore,
$$
r_2 = 2 - \sqrt{2} - r_1 \leq \frac{1}{2},
$$

which implies $\frac{3}{2} - \sqrt{2} \leq r_1 \leq \frac{1}{2}$.
Since the sum of the areas of $\odot O_1$ and $\odot O_2$ is $S = \pi (r_1^2 + r_2^2)$, we have
$$
\begin{array}{l}
\frac{S}{\pi} = r_1^2 + (2 - \sqrt{2} - r_1)^2 \\
= 2r_1^2 - 2(2 - \sqrt{2}) r_1 + 6 - 4\sqrt{2} \\
= 2\left(r_1 - \frac{2 - \sqrt{2}}{2}\right)^2 + 3 - 2\sqrt{2}.
\end{array}
$$

Since $\frac{3}{2} - \sqrt{2} \leq \frac{2 - \sqrt{2}}{2} \leq \frac{1}{2}$, when $r_1 = \frac{2 - \sqrt{2}}{2}$, $\odot O_1$ and $\odot O_2$ are equal circles, and the minimum value of their combined area is $(3 - 2\sqrt{2}) \pi$.
(2) As shown in Figure 16, draw the auxiliary line to form the right triangle $\triangle O_1O_2P$. Then
$$
\begin{array}{l}
O_1O_2 = r_1 + r_2, \\
O_1P = AB - r_1 - r_2 \\
= \frac{3}{2} - r_1 - r_2, \\
O_2P = BC - r_1 - r_2 \\
= 1 - r_1 - r_2.
\end{array}
$$

In the right triangle $\triangle O_1O_2P$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
O_1O_2^2 = O_1P^2 + O_2P^2 \\
\Rightarrow (r_1 + r_2)^2 \\
\quad = \left(\frac{3}{2} - r_1 - r_2\right)^2 + \left(1 - r_1 - r_2\right)^2 \\
\Rightarrow (r_1 + r_2)^2 - 5(r_1 + r_2) + \frac{13}{4} = 0 \\
\Rightarrow r_1 + r_2 = \frac{5}{2} + \sqrt{3} \text{ or } \frac{5}{2} - \sqrt{3}.
\end{array}
$$

Since $r_1 + r_2 < 1 + \frac{3}{2} = \frac{5}{2}$, we have
$$
r_1 + r_2 = \frac{5}{2} + \sqrt{3},
$$

which is not feasible, so we discard it.
Thus, $r_1 + r_2 = \frac{5}{2} - \sqrt{3}$.
It is easy to see that the sum of the areas of $\odot O_1$ and $\odot O_2$ is
$$
S = \pi (r_1^2 + r_2^2).
$$

And $r_1^2 + r_2^2 \geq \frac{(r_1 + r_2)^2}{2}$, with equality holding if and only if $r_1 = r_2$.

Therefore, when $r_1 = r_2$, the sum of the areas of $\odot O_1$ and $\odot O_2$ has a minimum value, and the minimum value is $\frac{\left(\frac{5}{2} - \sqrt{3}\right)^2 \pi}{2}$, which is $\left(\frac{37}{8} - \frac{5\sqrt{3}}{2}\right) \pi$.