A geometric progression has its first term a and common ratio r. Given that U1 + U3 = 40 and U4/U6 = 1/4. Find the common ratio and the fifth term.

$$r = \pm\sqrt{4}$$ $$r = \pm 2$$

The common ratio is $\boxed{\pm 2}$.

ii. Fifth term

If $r=2$:
$a(1 + 2^2) = 40$
$a(1 + 4) = 40$
$5a = 40$
$a = 8$

If $r=-2$:
$a(1 + (-2)^2) = 40$
$a(1 + 4) = 40$
$5a = 40$
$a = 8$
In both cases, the first term $a=8$.

Using $a=8$ and $r=2$:

$$U_5 = 8 \times (2)^4 = 8 \times 16 = 128$$

Using $a=8$ and $r=-2$:

$$U_5 = 8 \times (-2)^4 = 8 \times 16 = 128$$

The fifth term is $\boxed{128}$.

Part 11(b): Work Rate Problem

Let the total work be $W$.
A can finish the work in 10 days, so A's rate is $\frac{W}{10}$ per day.
B can finish the work in 15 days, so B's rate is $\frac{W}{15}$ per day.
C can finish the work in 30 days, so C's rate is $\frac{W}{30}$ per day.

Combined rate $R_{total} = \frac{W}{10} + \frac{W}{15} + \frac{W}{30}$

$$R_{total} = W \left(\frac{1}{10} + \frac{1}{15} + \frac{1}{30}\right)$$

$$R_{total} = W \left(\frac{3}{30} + \frac{2}{30} + \frac{1}{30}\right)$$ $$R_{total} = W \left(\frac{3+2+1}{30}\right)$$ $$R_{total} = W \left(\frac{6}{30}\right)$$ $$R_{total} = W \left(\frac{1}{5}\right)$$

Time = $\frac{\text{Total Work}}{\text{Combined Rate}}$

$$T = \frac{W}{W \left(\frac{1}{5}\right)}$$ $$T = \frac{1}{\frac{1}{5}}$$ $$T = 5 \text{ days}$$

The number of days required if A, B, and C work together is $\boxed{5 \text{ days}}$.

Part 12(a): Geometric Construction

i. Construct $\triangle PQR$ in which $|QR|=7\text{cm}$, $|PQ|=8\text{cm}$ and $\angle PQR=75^\circ$.

1. Draw a line segment $QR$ of length 7 cm.
2. At point $Q$, construct an angle of $75^\circ$. (This can be done by constructing a $90^\circ$ angle and a $60^\circ$ angle, then bisecting the angle between them).
3. Along the arm of the $75^\circ$ angle from $Q$, measure 8 cm and mark point $P$.
4. Join $P$ to $R$ to complete $\triangle PQR$.

ii. The Locus $L_1$ of points equidistant from $PR$ and $QR$.
This locus is the angle bisector of $\angle PRQ$.

1. Place the compass at point $R$ and draw an arc that intersects both line segments $PR$ and $QR$.
2. From the two intersection points on $PR$ and $QR$, draw two new arcs that intersect each other inside $\angle PRQ$.
3. Draw a straight line from $R$ through the intersection point of these two arcs. This line is $L_1$.

iii. The Locus $L_2$ of points equidistant from $Q$ and $R$.
This locus is the perpendicular bisector of the line segment $QR$.

1. Place the compass at point $Q$ and open it to a radius greater than half the length of $QR$. Draw an arc above $QR$ and an arc below $QR$.
2. With the same compass opening, place the compass at point $R$ and draw arcs above and below $QR$ that intersect the previously drawn arcs.
3. Draw a straight line through the two points where the arcs intersect. This line is $L_2$.

Part 12(b): Locate the point K, the intersection of $L_1$ and $L_2$.
The point $K$ is where the line $L_1$ (angle bisector of $\angle PRQ$) and the line $L_2$ (perpendicular bisector of $QR$) intersect. Mark this intersection point as $K$.

Part 12(c): Measure...
The question is incomplete, so this part cannot be answered.