If each block costs GH¢ 1.80 and the cost of labour is one-third the cost of a block, find the total cost of laying the blocks.

$$\text{Cost of labour per block} = \frac{1}{3} \times \text{Cost of one block}$$ $$\text{Cost of labour per block} = \frac{1}{3} \times \text{GHC } 1.80$$ $$\text{Cost of labour per block} = \text{GHC } 0.60$$

$$\text{Total cost per block laid} = \text{Cost of one block} + \text{Cost of labour per block}$$ $$\text{Total cost per block laid} = \text{GHC } 1.80 + \text{GHC } 0.60$$ $$\text{Total cost per block laid} = \text{GHC } 2.40$$

Assuming the question asks for the cost per block laid, the total cost is $\boxed{\text{GHC 2.40}}$.

Geometry Question:
In the diagram, MNPQ is a circle with centre O. $\angle MNQ = y^\circ$, $\angle NEP = x^\circ$, $\angle NMP = (100-x)^\circ$ and $\angle NFP = (5x)^\circ$. Find the values of $x$ and $y$.

The measure of an inscribed angle is half the measure of its intercepted arc.
For $\angle MNQ$:

$$\angle MNQ = y^\circ \implies \text{arc MQ} = 2 \times \angle MNQ = 2y^\circ$$

For $\angle NMP$:

$$\angle NMP = (100-x)^\circ \implies \text{arc NP} = 2 \times \angle NMP = 2(100-x)^\circ$$

The angle formed by two chords intersecting inside a circle is half the sum of the measures of the intercepted arcs.

$$\angle NFP = \frac{1}{2} (\text{arc NP} + \text{arc MQ})$$

Given $\angle NFP = (5x)^\circ$:

$$5x = \frac{1}{2} (2(100-x) + 2y)$$ $$5x = (100-x) + y$$ $$5x = 100 - x + y$$ $$6x - y = 100 \quad (\text{Equation 1})$$

The angle formed by two secants intersecting outside a circle is half the difference of the measures of the intercepted arcs.

$$\angle NEP = \frac{1}{2} (\text{arc MQ} - \text{arc NP})$$

Given $\angle NEP = x^\circ$:

$$x = \frac{1}{2} (2y - 2(100-x))$$ $$x = y - (100-x)$$ $$x = y - 100 + x$$ $$0 = y - 100$$ $$y = 100$$

Substitute $y=100$ into Equation 1:

$$6x - 100 = 100$$ $$6x = 100 + 100$$ $$6x = 200$$ $$x = \frac{200}{6}$$ $$x = \frac{100}{3}$$ $$x \approx 33.33$$

The values are $x = \frac{100}{3}$ and $y = 100$.
The values of $x$ and $y$ are $\boxed{x = \frac{100}{3} \text{ or } 33.3^\circ \text{ and } y = 100^\circ}$.