SECTION B: Answer Any five (5) Questions

The number "25" in base x is equivalent to $2x^1 + 5x^0 = 2x + 5$ in base 10.
The number "52" in base x is equivalent to $5x^1 + 2x^0 = 5x + 2$ in base 10.

According to the problem, twice the first number equals the second number:

Subtract $4x$ from both sides:

Subtract 2 from both sides:

The value of x is $\boxed{\textbf{8}}$.

2. Evaluate each of the following without using tables or calculator, to four significant figures.
Given $\sqrt{5} = 2.2361$ and $\sqrt{7} = 2.6458$.

i) $\sqrt{72} - \sqrt{7}$

We need to evaluate $\sqrt{72} - \sqrt{7}$.
We can simplify $\sqrt{72}$ as $\sqrt{36 \times 2} = 6\sqrt{2}$.
The expression becomes $6\sqrt{2} - \sqrt{7}$.
However, the problem only provides values for $\sqrt{5}$ and $\sqrt{7}$, and explicitly states "without using tables or calculator". Since $\sqrt{2}$ is not provided, and we cannot use a calculator to find its value, this expression cannot be evaluated to four significant figures using the given information.

This expression cannot be evaluated to four significant figures with the provided values for $\sqrt{5}$ and $\sqrt{7}$ only, as $\sqrt{2}$ is required and not given.

ii) $\frac{\sqrt{5} + 2\sqrt{5}}{\sqrt{5}}$

To four significant figures, the value is $\boxed{\textbf{3.000}}$.

3. Evaluate $\log_2 \sqrt{\frac{a^2 b}{c^2}}$
Given $\log_2 a = 3$, $\log_2 b = -2$, and $\log_2 c = 5$.

Using the power rule for logarithms, $\log x^n = n \log x$:

Using the quotient rule, $\log \frac{x}{y} = \log x - \log y$:

Using the product rule, $\log (xy) = \log x + \log y$, and the power rule again:

Substitute $\log_2 a = 3$, $\log_2 b = -2$, and $\log_2 c = 5$:

The value of the expression is $\boxed{\textbf{-3}}$.

4. Find the mean of the numbers $5 + x_1, 5 + x_2, 5 + x_3, 5 + x_4$.
Given that the mean of $x_1, x_2, x_3, x_4$ is 7.

The mean of $x_1, x_2, x_3, x_4$ is 7. There are 4 numbers.

The second set of numbers is $5 + x_1, 5 + x_2, 5 + x_3, 5 + x_4$.
Their sum is:

There are 4 numbers in the second set.

The mean of the numbers is $\boxed{\textbf{12}}$.

5. Calculate the mean age of the class.
The frequency table is:

The mean age of the class is $\boxed{\textbf{13 years}}$.

6. Make $u$ the subject of the relation $fx = w\left(\frac{u^2 - v^2}{2a}\right)$.

Multiply both sides by $2a$:

Divide both sides by $w$:

Add $v^2$ to both sides:

Take the square root of both sides:

The subject of the relation is \boxed{\textbf{u = \pm \sqrt{\frac{2afx}{w} + v^2}}}.

7. Calculate the:

i) Distance XZ

ii) Bearing of Z from X

Draw a diagram with North lines.
The bearing of Y from X is $030^\circ$.
The bearing of Z from Y is $078^\circ$.
Draw a North line at Y. The angle between the North line at Y and the line YX (back towards X) is $30^\circ$ (alternate interior angles with the bearing from X to Y).
The angle between the North line at Y and the line YZ is $78^\circ$.
Both angles are measured clockwise from the North line at Y.
Therefore, the interior angle $\angle XYZ = 78^\circ - 30^\circ = 48^\circ$.

In $\triangle XYZ$, we have $XY = 10$ km, $YZ = 15$ km, and $\angle XYZ = 48^\circ$.
Using the Cosine Rule, $XZ^2 = XY^2 + YZ^2 - 2(XY)(YZ)\cos(\angle XYZ)$:

Rounding to one decimal place, the distance XZ is $\boxed{\textbf{11.1 km}}$.

Using the Sine Rule, $\frac{\sin(\angle YXZ)}{YZ} = \frac{\sin(\angle XYZ)}{XZ}$:

The bearing of Y from X is $030^\circ$.
The bearing of Z from X is the bearing of Y from X plus the angle $\angle YXZ$.

The bearing of Z from X is \boxed{\textbf{119.4^\circ}}.