Determine the diameter of a hockey ball.

The radius of a hockey ball is $35.5 \text{ mm}$.
The diameter of a hockey ball is $D_{ball} = 2 \times 35.5 \text{ mm} = 71 \text{ mm}$.

The diameter of Container A is $73 \text{ mm}$.
Number of balls across diameter = $\frac{\text{Container A diameter}}{\text{Hockey ball diameter}} = \frac{73 \text{ mm}}{71 \text{ mm}} \approx 1.02$.
Since only whole balls can fit, 1 hockey ball can fit across the diameter.

The height of Container A is $280 \text{ mm}$.
Number of balls along height = $\frac{\text{Container A height}}{\text{Hockey ball diameter}} = \frac{280 \text{ mm}}{71 \text{ mm}} \approx 3.94$.
Since only whole balls can fit, 3 hockey balls can fit along the height.

Total balls = (balls across diameter) $\times$ (balls along height)
Total balls = $1 \times 3 = 3$
Thus, $\boxed{\text{3 hockey balls}}$ can fit into CONTAINER A.

1.3.2

The diameter of a hockey ball is $71 \text{ mm}$ (from 1.3.1).

Length = $240 \text{ mm}$.
Width = $17 \text{ cm} = 17 \times 10 \text{ mm} = 170 \text{ mm}$.
Height = $100 \text{ mm}$.

Number of balls along length = $\frac{240 \text{ mm}}{71 \text{ mm}} \approx 3.38$. So, 3 balls.
Number of balls along width = $\frac{170 \text{ mm}}{71 \text{ mm}} \approx 2.39$. So, 2 balls.
Number of balls along height = $\frac{100 \text{ mm}}{71 \text{ mm}} \approx 1.40$. So, 1 ball.

Total balls = (balls along length) $\times$ (balls along width) $\times$ (balls along height)
Total balls = $3 \times 2 \times 1 = 6$
Thus, $\boxed{\text{6 hockey balls}}$ can fit into CONTAINER B.

1.3.3

Container A is a cylindrical shape, which is a better match for the spherical hockey balls. This allows for a more snug fit, minimizing wasted space and preventing the balls from moving around and potentially getting damaged during transport.