Determine the interior angle ABC.

The bearing of B from A is $053^\circ$. This means the angle from the North line at A to AB is $53^\circ$.
The bearing of C from B is $161^\circ$. This means the angle from the North line at B to BC is $161^\circ$.
Since the North lines at A and B are parallel, the alternate interior angle to $53^\circ$ (angle NAB) is the angle between the line BA and the North line at B, which is $53^\circ$.
Therefore, $\angle ABC = 161^\circ - 53^\circ = 108^\circ$.

Given $AB = 15\, \text{m}$, $BC = 18\, \text{m}$, and $\angle ABC = 108^\circ$.

Correct to three significant figures, the length of AC is $\boxed{26.8\, \text{m}}$.

The bearing of B from A is $053^\circ$. The bearing of C from A is the bearing of B from A plus $\angle BAC$.

Correct to three significant figures, the bearing of C from A is $\boxed{092.7^\circ}$.

The pole BT is vertical, so $\angle ABT = 90^\circ$.
Given $AB = 15\, \text{m}$ and $\angle ATB = 58^\circ$.

Correct to three significant figures, the height of the pole BT is $\boxed{9.37\, \text{m}}$.