At $09:00$, hour hand and minute hand are at right angles to each other. At all the other instances, the minute hand has moved, so the hour hand would have moved too. Hence, the angle formed will not be ${90}^{\circ }$ in those cases.

If you draw everything according to the given question, you would get: 

We can see that PQ is parallel to CD or vice-versa.

$\angle XYZ$ is also known as a right angle if it is equal to ${90}^{\circ }$.

If you follow all the steps, you would get this:

If you measure $\angle PAB$, the value which you would get is ${60}^{\circ }$.

Draw the circle with center O using compass and ruler. Now, since smaller circles are non intersecting and passing through the point O, there is only one possibility.

After joining PQ, drawing diameter RS perpendicular to PQ and joining PQRS, you will get:

PQRS is a rhombus.

At $09:00$, hour hand and minute hand are at right angles to each other. At all the other instances, the minute hand has moved, so the hour hand would have moved too. Hence, the angle formed will not be ${90}^{\circ }$ in those cases.

Line segments are represented with bar on top of their name, which contains two letters which are the two ends.

If you follow all the steps, you would get this:

If you measure $\angle PAB$, the value which you would get is ${60}^{\circ }$.

We know that all squares are rhombuses and parallelograms. We can also make a pentagon as shown in the figure. But we can never get a trapezium using $2$ set squares with angles (${45}^{\circ }$-${45}^{\circ }$-${90}^{\circ }$).

We can draw ${60}^{\circ }$ and bisect it to get ${30}^{\circ }.$
Similarly, we can bisect ${30}^{\circ }$ and get ${15}^{\circ }.$
Also, we can draw ${90}^{\circ }$ and bisect it to get ${45}^{\circ }.$
But, ${50}^{\circ }$ cannot be drawn using compass and ruler to get its bisected angle of ${25}^{\circ }$.