Identify the properties of the equilateral triangle PQR.

Here are the solutions to the problems:

1.a. Calculate the area of the shaded region.

Step 1: Identify the properties of the equilateral triangle PQR.
The side length of the equilateral triangle PQR is $s = 18$ cm.
All angles in an equilateral triangle are $60^\circ$, so $\angle P = \angle Q = \angle R = 60^\circ$.

Step 2: Calculate the area of the equilateral triangle PQR.
The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$.

$$\text{Area of } \triangle PQR = \frac{\sqrt{3}}{4} (18)^2 = \frac{\sqrt{3}}{4} \times 324 = 81\sqrt{3} \text{ cm}^2$$

Using $\sqrt{3} \approx 1.73205$:

$$\text{Area of } \triangle PQR \approx 81 \times 1.73205 = 140.29605 \text{ cm}^2$$

Step 3: Determine the radius of the sector PAB.
The arc is centered at P and touches QR at M. This means PM is the radius of the sector.
In an equilateral triangle, the line segment from a vertex to the midpoint of the opposite side (PM) is also an altitude.
Consider the right-angled triangle PQM, where $PQ = 18$ cm and $QM = \frac{1}{2} QR = \frac{1}{2} \times 18 = 9$ cm.
Using the Pythagorean theorem: $PM^2 + QM^2 = PQ^2$

$$PM^2 + 9^2 = 18^2$$ $$PM^2 + 81 = 324$$ $$PM^2 = 324 - 81 = 243$$ $$PM = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3} \text{ cm}$$

So, the radius of the sector is $r = 9\sqrt{3}$ cm.

Step 4: Calculate the area of the sector PAB.
The angle of the sector is $\angle QPR = 60^\circ$.
The formula for the area of a sector is $\frac{\theta}{360^\circ} \pi r^2$.
Using $\pi = \frac{22}{7}$:

$$\text{Area of sector PAB} = \frac{60}{360} \times \frac{22}{7} \times (9\sqrt{3})^2$$ $$= \frac{1}{6} \times \frac{22}{7} \times (81 \times 3)$$ $$= \frac{1}{6} \times \frac{22}{7} \times 243$$ $$= \frac{11}{3 \times 7} \times 243$$ $$= \frac{11}{21} \times 243$$ $$= \frac{11 \times 81}{7} = \frac{891}{7} \text{ cm}^2$$ $$\text{Area of sector PAB} \approx 127.28571 \text{ cm}^2$$

Step 5: Calculate the area of the shaded region.
The shaded region is the area of the triangle PQR minus the area of the sector PAB.

$$\text{Area of shaded region} = \text{Area of } \triangle PQR - \text{Area of sector PAB}$$ $$\text{Area of shaded region} \approx 140.29605 - 127.28571$$ $$\text{Area of shaded region} \approx 13.01034 \text{ cm}^2$$

Correct to two decimal places:
The area of the shaded region is $\boxed{13.01 \text{ cm}^2}$.

1.b. Solve the equation $a - \sqrt{a^2 + 12} + 2 = 0$

Step 1: Isolate the square root term.

$$a + 2 = \sqrt{a^2 + 12}$$

Step 2: Square both sides of the equation to eliminate the square root.

$$(a+2)^2 = (\sqrt{a^2 + 12})^2$$ $$a^2 + 4a + 4 = a^2 + 12$$

Step 3: Solve for $a$.
Subtract $a^2$ from both sides:

$$4a + 4 = 12$$

Subtract 4 from both sides:

$$4a = 12 - 4$$ $$4a = 8$$

Divide by 4:

$$a = \frac{8}{4}$$ $$a = 2$$

Step 4: Check the solution by substituting $a=2$ back into the original equation.

$$2 - \sqrt{2^2 + 12} + 2 = 0$$ $$2 - \sqrt{4 + 12} + 2 = 0$$ $$2 - \sqrt{16} + 2 = 0$$ $$2 - 4 + 2 = 0$$ $$0 = 0$$

The solution is correct.
The value of a is $\boxed{2}$.

1.c. The ratio of the age of a mother to that of the daughter is 11:3. If the mother is 55 years old, what will be the ratio of their ages in 5 years time?

Step 1: Determine the current ages of the mother and daughter.
Let the mother's age be $M$ and the daughter's age be $D$.
Given ratio $M:D = 11:3$.
Given mother's current age $M = 55$ years.
We can write the ratio as $\frac{M}{D} = \frac{11}{3}$.
Substitute $M=55$:

$$\frac{55}{D} = \frac{11}{3}$$

To find $D$, cross-multiply:

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