(1) 当点 $M$ 与点 $B$ 重合时, 求 $t$ 的值;
(2) 当 $t$ 为何值时, $\triangle A P Q$ 与 $\triangle B M F$ 全等;
(3) 求 $S$ 与 $t$ 的函数关系式;
(4) 以线段 $P Q$ 为边, 在 $P Q$ 右侧作等边三角形 $P Q E$, 当 $2 \leqslant t \leqslant 4$ 时, 请直接写出点 $E$ 运动路径的长.

\begin{aligned} & \because \angle A=60^{\circ}, \\ & \therefore P A=\frac{1}{2} A B=2, \\ & \therefore t=2 \\ \end{aligned}

(2)① 当 $0 \leqslant 1 \leqslant 2$ II的,
\begin{aligned} & \because A M=2 t, \quad \therefore B M=4-2 t, \\ & \because \triangle A P Q \cong \triangle B M F, \therefore A P=B M, \\ & \therefore t=4-2 t, \\ & \therefore t=\frac{4}{3}, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \cdots \end{aligned}
②当 $2 < 1 \leqslant 4$ 时,
\begin{aligned} & \because A M=2 t, \\ & \therefore B M=2 t-4, \\ & \because \triangle A P Q \cong \triangle B M F, \therefore A P=B M, \\ & \therefore t=2 t-4, \end{aligned}

\begin{aligned} & \therefore t=4 . \\ & \therefore t=4 \text { 或 } t=\frac{4}{3} . \end{aligned}

(3) ① 当 $0 \leqslant t \leqslant 2$ 时, 如图 ,
\begin{aligned} & P Q=\frac{\sqrt{3}}{2} t, \\ & \therefore M Q=\frac{3}{2} t, \\ & \therefore S=S_{\triangle P Q M}=\frac{3 \sqrt{3}}{8} t^2 . \end{aligned}

②当 $2 < 1 \leqslant 4$ 时, 如图 ,
\begin{aligned} & \because B F=t-2, \quad M F=\sqrt{3}(t-2), \\ & \therefore S_{\triangle B F M}=\frac{\sqrt{3}}{2}(t-2)^2, \\ & \therefore S=S_{\triangle P Q M}-S_{\triangle B F M}=-\frac{\sqrt{3}}{8} t^2+2 \sqrt{3} t-2 \sqrt{3}, \\ & \therefore S=\left\{\begin{array}{l} \frac{3 \sqrt{3}}{8} t^2(0 \leq t \leq 2), \\ -\frac{\sqrt{3}}{8} t^2+2 \sqrt{3} t-2 \sqrt{3}(2 < t \leq 4) . \end{array}\right. \end{aligned}

(4) $E$ 的运动路径长为 $2 \sqrt{7}-\sqrt{7}=\sqrt{7}$.

$\because \triangle P Q E$ 为正三角形,
$$\therefore P E=\frac{\sqrt{3}}{2} t \text {, }$$

$\therefore \angle P A E$ 为定值.
$\therefore E$ 的运动轨迋为直线,
\begin{aligned} & A E=\sqrt{A P^2+P E^2}=\frac{\sqrt{7}}{2} t, \\ & \text { 当 } t=2 \text { 时, } A E=\sqrt{7}, \\ & \text { 当 } t=4 \text { 时, } A E=2 \sqrt{7}, \end{aligned}
$\therefore E$ 的运动路径长为 $2 \sqrt{7}-\sqrt{7}=\sqrt{7}$.