Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

The projection of line VP onto the plane PQRS is the line segment connecting P to the foot of the perpendicular from V to the plane, which is O. So, the projection is OP.

In the rectangular base PQRS, consider the right-angled triangle PQR.
Using the Pythagorean theorem:

Since O is the center of the rectangle, it is the midpoint of the diagonal PR.

The length of the projection of line VP on the plane PQRS is $\boxed{\text{10 cm}}$.

The angle between line VP and the plane PQRS is the angle between VP and its projection OP. This is $\angle VPO$.

Triangle VOP is a right-angled triangle with the right angle at O.
We know $VP = 18$ cm (hypotenuse) and $OP = 10$ cm (adjacent to $\angle VPO$).

The size of the angle between line VP and the plane PQRS is $\boxed{56.3^\circ}$ (to one decimal place).

The line of intersection of plane VQR and plane PQRS is the line segment QR.
Let M be the midpoint of QR.
Since triangle VQR is an isosceles triangle ($VQ = VR = 18$ cm), the line segment VM from the vertex V to the midpoint M of the base QR is perpendicular to QR. So, $VM \perp QR$.
In the rectangular base PQRS, the line segment OM (connecting the center O to the midpoint M of QR) is perpendicular to QR. So, $OM \perp QR$.
Therefore, the angle between the planes VQR and PQRS is $\angle VMO$.

Since O is the center of the rectangle and M is the midpoint of QR, OM is parallel to PQ and its length is half of PQ.

In the right-angled triangle VOP (from part b), we can find VO using the Pythagorean theorem:

Triangle VOM is a right-angled triangle with the right angle at O.
We know $VO = 4\sqrt{14}$ cm (opposite to $\angle VMO$) and $OM = 8$ cm (adjacent to $\angle VMO$).

The size of the angle between the planes VQR and PQRS is $\boxed{61.9^\circ}$ (to one decimal place).