A) Find the coordinates of D. b) Find the value of x in the diagram.

$$M_{AC} = \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right)$$ $$M_{AC} = \left(\frac{1 + (-1)}{2}, \frac{2 + (-1)}{2}\right)$$ $$M_{AC} = \left(\frac{0}{2}, \frac{1}{2}\right)$$ $$M_{AC} = \left(0, \frac{1}{2}\right)$$

$$M_{BD} = \left(\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}\right)$$ $$\left(\frac{3 + x_D}{2}, \frac{-1 + y_D}{2}\right) = \left(0, \frac{1}{2}\right)$$

Equating the x-coordinates:

$$\frac{3 + x_D}{2} = 0$$ $$3 + x_D = 0$$ $$x_D = -3$$

Equating the y-coordinates:

$$\frac{-1 + y_D}{2} = \frac{1}{2}$$ $$-1 + y_D = 1$$ $$y_D = 1 + 1$$ $$y_D = 2$$

The coordinates of D are $\boxed{(-3, 2)}$.

The exterior angle at P is $3x - 18^\circ$.
The remote interior angle at Q is $52^\circ$.
The remote interior angle at R is $2x + 10^\circ$.

$$3x - 18 = 52 + (2x + 10)$$

$$3x - 2x = 62 + 18$$ $$x = 80$$

The value of x is $\boxed{80}$.

$$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$$ $$3\frac{1}{3} = \frac{3 \times 3 + 1}{3} = \frac{10}{3}$$

The expression becomes:

$$\left(\frac{3}{5} + \frac{3}{2}\right) \div \frac{1}{4} \times \frac{10}{3}$$

Find a common denominator for $\frac{3}{5}$ and $\frac{3}{2}$, which is 10.

$$\frac{3}{5} + \frac{3}{2} = \frac{3 \times 2}{5 \times 2} + \frac{3 \times 5}{2 \times 5} = \frac{6}{10} + \frac{15}{10} = \frac{21}{10}$$

The expression is now:

$$\frac{21}{10} \div \frac{1}{4} \times \frac{10}{3}$$

First, the division. To divide by a fraction, multiply by its reciprocal.

$$\frac{21}{10} \div \frac{1}{4} = \frac{21}{10} \times \frac{4}{1} = \frac{84}{10} = \frac{42}{5}$$

The expression is now:

$$\frac{42}{5} \times \frac{10}{3}$$

$$\frac{420}{15} = \frac{84}{3} = 28$$

The simplified value is $\boxed{28}$.

$$m = \frac{a - t}{tb + 1}$$

The formula with m as the subject is $\boxed{m = \frac{a - t}{tb + 1}}$.