5. Tangram is a familiar puzzle toy. Now, please use the cards provided to make a set of tangrams as shown in Figure 1, and take the four pieces numbered 1 to 4 to make a four-piece tangram.
(1) Use the four pieces of the four-piece tangram to form the step shape shown in Figure 2. Draw the assembly method directly on Figure 2 with thick lines.
(2) Figure 3 shows a four-level step of a cube, where the length, width, and height of each step are equal. It is known that the height $\mathrm{AD}=h=3$ cm, the width $\mathrm{DE}=b=1$ cm, and the length $\mathrm{AC}=a=\mathbf{8}$ cm. A clever mouse crawls from point A to point B along the surface of the steps (assuming it can crawl vertically on vertical surfaces), and it takes the shortest path. There is also a smart mousetrap that can be placed at any point on DE, FG, or HI. If it is placed on DE, then its movement path must be perpendicular to DE. Similarly, if it is placed on FG or HI, then its movement path must be perpendicular to FG or HI. It is known that the mouse and the smart mousetrap start at the same time, and the speed of the mouse $v=17$ cm/s. Prove that to catch the mouse exactly, the speed of the smart mousetrap is independent of its placement position, and find its speed.
【Analysis】As shown in the figure, if the steps are stretched into a plane figure, then the shortest path is the line connecting $\mathrm{AB}$. Clearly, $\mathrm{AC}=\mathrm{BJ}=8, \mathrm{AJ}=15$, so, by the Pythagorean theorem, $\mathrm{AB}=17$.
Next, we prove that the speed of the smart mousetrap placed at any point on $\mathrm{AJ}$ is the same when it captures the mouse. Take any point $\mathrm{P}$ on $\mathrm{AJ}$, and draw a perpendicular line from $\mathrm{P}$ to $\mathrm{AJ}$ intersecting $\mathrm{AB}$ at $\mathrm{Q}$.
Clearly, the mousetrap captures the mouse exactly at point $\mathrm{Q}$.
Clearly, $\mathrm{PQ} \parallel \mathrm{BJ}$, so $\frac{P Q}{B J}=\frac{A Q}{A B} \Rightarrow P Q=\frac{A Q \times 8}{17}$,
and the time it takes for the mouse to run to point $\mathrm{Q}$ is $A Q \div 17=\frac{A Q}{17}$.
Therefore, the speed of the mousetrap should be $P Q \div \frac{A Q}{17}=\frac{A Q \times 8}{17} \times \frac{17}{A Q}=8$.
Clearly, the speed of the mousetrap is independent of its position and should be 8 cm/s.