(1) 求 $m$ 的值;
(2) 若 $P A=2 A B$, 求 $k$ 的值.

$\therefore$ 代入得 $m=\frac{4}{1}=4$, $\therefore m=4 ;$ (2) 令 $y=0$, 即 $k x+b=0$, $\therefore x=-\frac{b}{k}, A\left(-\frac{b}{k}, 0\right)$, 令 $x=0, y=b$, $\therefore B(0, b)$, $\because P A=2 A B$,

(1) $B$ 在 $y$ 轴正半轴时, $b > 0$,
$$\because P A=2 A B,$$

\begin{aligned} &\text { 又 } B_{1} O \perp A_{1} H, \angle P A_{1} O=\angle B_{1} A_{1} O, \\ &\therefore \triangle A_{1} O B_{1} \backsim \triangle A_{1} H P, \end{aligned}
\begin{aligned} &\therefore \frac{A_{1} B_{1}}{A_{1} P}=\frac{A_{1} O}{A_{1} H}=\frac{B_{1} O}{P H}=\frac{1}{2}, \\ &\therefore B_{1} O=\frac{1}{2} P H=4 \times \frac{1}{2}=2, \\ &\therefore b=2, \\ &\therefore A_{1} O=O H=1, \\ &\therefore\left|-\frac{b}{k}\right|=1, \\ &\therefore k=2 ; \end{aligned}
(2) $B$ 在 $y$ 轴负半轴时, $b < 0, \quad$ 过 $P$ 作 $P Q \perp y$ 轴,
$\because P Q \perp B_{2} Q, A_{2} O \perp B_{2} Q, \angle A_{2} B_{2} O=\angle A B_{2} Q$, $\therefore \triangle A_{2} O B_{2} \triangle P Q B_{2}$, $\therefore \frac{A_{2} B_{2}}{P B_{2}}=\frac{1}{3}=\frac{A_{2} O}{P Q}=\frac{B_{2} O}{B_{2} Q}$, $\therefore A O=\left|-\frac{b}{k}\right|=\frac{1}{3} P O=\frac{1}{3}, B_{2} O=\frac{1}{3} B_{2} Q=\frac{1}{2} O Q=|b|=2$, $\therefore b=-2$, $\therefore k=6$, 综上, $k=2$ 或 $k=6 .$