答案：
解: (1) ① $\because$ 点 $D(12,3)$ 在直线 $y = \frac{1}{3}x + b$ 上, $\therefore 3 = \frac{1}{3}×12 + b$, $\therefore b = -1$, $\therefore$ 直线 $l$ 的函数表达式为 $y = \frac{1}{3}x - 1$, $\therefore C(3,0)$, $\because DE \perp y$ 轴, $\therefore OE = 3$, $\because CA \perp DE$, $\therefore AC = OE = 3$, $\therefore DB = AC = 3$, $\therefore B(9,3)$. 设直线 $BC$ 的函数表达式为 $y = kx + a$, 则有 $\begin{cases}9k + a = 3,\\3k + a = 0,\end{cases}$ 解得 $\begin{cases}k = \frac{1}{2},\\a = -\frac{3}{2},\end{cases}$ $\therefore$ 直线 $BC$ 的函数表达式为 $y = \frac{1}{2}x - \frac{3}{2}$. ② 设 $P(m,\frac{1}{2}m - \frac{3}{2})$, 则 $R(m,0)$, $Q(m,\frac{1}{3}m - 1)$, $S(m,3)$, $\because QS = QR$, $\therefore 3 - (\frac{1}{3}m - 1) = \frac{1}{3}m - 1$, $\therefore m = \frac{15}{2}$, $\therefore P(\frac{15}{2},\frac{9}{4})$.
(2) 结论: 不变. 如答图, 过点 $D$ 作 $DT \perp x$ 轴于点 $T$. 设 $D(d,\frac{1}{3}d + b)$,

$\because C(-3b,0)$, $\therefore OC = -3b$, $OT = d$, $DT = \frac{1}{3}d + b$, $\therefore CT = OT - OC = d + 3b$, $\because AC = DT = BD = \frac{1}{3}d + b$, $\therefore B(\frac{2}{3}d - b,\frac{1}{3}d + b)$, $\therefore$ 直线 $BC$ 的函数表达式为 $y = \frac{1}{2}x + \frac{3}{2}b$. 设 $P(t,\frac{1}{2}t + \frac{3}{2}b)$, 则 $R(t,0)$, $Q(t,\frac{1}{3}t + b)$, $\therefore PQ = \frac{1}{2}t + \frac{3}{2}b - (\frac{1}{3}t + b) = \frac{1}{6}t + \frac{1}{2}b$, $CR = t - (-3b) = t + 3b$, $\therefore \frac{PQ}{CR} = \frac{\frac{1}{6}t + \frac{1}{2}b}{t + 3b} = \frac{1}{6}$.

答案：$y = -2x + 6$

答案：$y = -\frac{1}{3}x + 2$

答案：解: (1) 将 $B(8,8)$ 代入 $y = kx$, 得 $8k = 8$, 则 $k = 1$.
(2) 当 $x = t$ 时, $y = \frac{1}{2}t + 4$, 即 $N(t,\frac{1}{2}t + 4)$; $y = t$, 即 $M(t,t)$. $\because 0 \leq t \leq 8$, $\therefore NM = \frac{1}{2}t + 4 - t = 4 - \frac{1}{2}t$, $\therefore S_{\triangle OMN} = \frac{1}{2}MN \cdot OP = \frac{1}{2}(4 - \frac{t}{2}) \cdot t = 2t - \frac{1}{4}t^2$.