18. (14 points) As shown in Figure 8, in the Cartesian coordinate system $x O y$, the rectangle $O A B C$ has its sides on the $x$-axis and $y$-axis, with $O A=10 \text{ cm}$ and $O C=6 \text{ cm}$. There are two moving points $P$ and $Q$ starting from $O$ and $A$ simultaneously. Point $P$ moves uniformly along $O A$, and point $Q$ moves uniformly along $A B$. It is known that the speed of point $P$ is $1 \text{ cm/s}$.
(1) Suppose the speed of point $Q$ is $\frac{1}{2} \text{ cm/s}$, and the movement time is $t \text{ s}$.
(1) When the area of $\triangle C P Q$ is minimized, find the coordinates of point $Q$;
(2) When $\triangle C O P$ and $\triangle P A Q$ are similar, find the coordinates of point $Q$.
(2) Suppose the speed of point $Q$ is $a \text{ cm/s}$. Is there a value of $a$ such that $\triangle O C P$ is similar to both $\triangle P A Q$ and $\triangle C B Q$? If it exists, find the value of $a$ and the coordinates of point $Q$ at that time; if not, explain the reason.
18. (1)(i) Notice that
$$
\begin{aligned}
& S_{\triangle C P Q}=S_{\text {quadrilateral OABC }}-S_{\triangle O C P}-S_{\triangle P R Q}-S_{\triangle B C C} \\
= & 60-\frac{1}{2} \times 6 t-\frac{1}{2}(10-t) \cdot \frac{1}{2} t- \\
& \frac{1}{2} \times 10\left(6-\frac{1}{2} t\right) \\
= & \frac{1}{4} t^{2}-3 t+30 \\
= & \frac{1}{4}(t-6)^{2}+21(0 \leqslant t \leqslant 10) .
\end{aligned}
$$

Therefore, when $t=6$, the minimum value of $S_{\triangle C P Q}$ is 21.
At this time, point $Q(10,3)$.
(ii) Let $\angle O P C=\angle 1, \angle A P Q=\angle 2$, $\angle P Q A=\angle 3, \angle B Q C=\angle 4, \angle Q C B=\angle 5$.
When $\angle 1=\angle 2$, we have
$$
\frac{O C}{O P}=\frac{Q A}{P A} \Rightarrow \frac{6}{t}=\frac{\frac{1}{2} t}{10-t} \Rightarrow \frac{1}{2} t^{2}+6 t-60=0
$$
$\Rightarrow t_{1}=-6+2 \sqrt{39}, t_{2}=-6-2 \sqrt{39}$ (discard).
When $\angle 1=\angle 3$, we have $\frac{6}{t}=\frac{10-t}{\frac{1}{2} t}$, solving gives $t=7$.
Therefore, when $t=-6+2 \sqrt{39}$ or 7, that is, when point $Q(10,-3+\sqrt{39})$ or $Q\left(10, \frac{7}{2}\right)$, $\triangle C O P$ and $\triangle P A Q$ are similar.
(2) Assume there exists a value of $a$ such that $\triangle O C P$ is similar to both $\triangle P A Q$ and $\triangle C B Q$. Let the time $P$ and $Q$ move be $t \mathrm{~s}$. Then $O P=t, A Q=a t$.
(i) When $\angle 1=\angle 3=\angle 4$,
$$
\frac{O C}{O P}=\frac{P A}{A Q}=\frac{B C}{B Q} \Rightarrow \frac{6}{t}=\frac{10-t}{a t}=\frac{10}{6-a t} \text {. }
$$

Solving gives $t_{1}=2, t_{2}=18$ (discard).
At this time, $a=\frac{4}{3}$, point $Q\left(10, \frac{8}{3}\right)$.
(ii) When $\angle 1=\angle 3=\angle 5$, $\angle C P Q=$ $\angle C Q P=90^{\circ}$ does not hold.
(iii) When $\angle 1=\angle 2=\angle 4$,
$$
\begin{array}{l}
\frac{O C}{O P}=\frac{Q A}{A P}=\frac{B C}{B Q} \\
\Rightarrow \frac{6}{t}=\frac{a t}{10-t}=\frac{10}{6-a t} \\
\Rightarrow\left\{\begin{array}{l}
60-6 t=a t^{2}, \\
36-6 a t=10 t,
\end{array}\right.
\end{array}
$$

From equation (2), we get $a t=\frac{36-10 t}{6}$.
Substituting into equation (1) and rearranging gives $5 t^{2}-36 t+180=0$. Since $\Delta\angle 5$.
Therefore, there does not exist such a value of $a$.
In summary, there exists a value of $a$ such that $\triangle O C P$ is similar to both $\triangle P A Q$ and $\triangle C B Q$.
At this time, $a=\frac{4}{3}$, point $Q\left(10, \frac{8}{3}\right)$.