Show that the volume, V cm³, of liquid in the container when the depth is x cm is given by V = (1/12)πx³. Find the exact rate at which the depth of the liquid is increasing at the instant when the depth is 8 cm.

From the diagram, for the liquid cone, the depth is $x$ and the radius of the surface is $r_L$. The angle $\alpha$ is between the axis and the slant height. In the right-angled triangle formed by the depth, radius, and slant height, we have:

We are given $\alpha = \tan^{-1}\left(\frac{1}{2}\right)$, which means $\tan\alpha = \frac{1}{2}$.
Substituting this into the equation from Step 1:

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. For the liquid, $r = r_L$ and $h = x$.

This shows the required formula.

We are given that liquid is added at a rate of $14 \text{ cm}^3$ per minute, so $\frac{dV}{dt} = 14 \text{ cm}^3/\text{min}$.
We need to find the rate at which the depth is increasing, which is $\frac{dx}{dt}$, when $x = 8 \text{ cm}$.

From part (i), $V = \frac{1}{12}\pi x^3$.

The chain rule states $\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}$.
Substitute the expressions from Step 1 and Step 2:

We need to find $\frac{dx}{dt}$ when $x = 8 \text{ cm}$.

The exact rate at which the depth of the liquid is increasing is $\boxed{\frac{7}{8\pi} \text{ cm/min}}$.