Let f(x) = (4x)/(x-1). (i) Show that f is injective. (ii) Find the composite function f(f(x)) and determine its domain.

Divide both sides by 4:

$$\frac{x_1}{x_1-1} = \frac{x_2}{x_2-1}$$

Cross-multiply:

$$x_1(x_2-1) = x_2(x_1-1)$$ $$x_1x_2 - x_1 = x_1x_2 - x_2$$

Subtract $x_1x_2$ from both sides:

$$-x_1 = -x_2$$

Multiply by -1:

$$x_1 = x_2$$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is injective.

4. (i) (b)

$$(f \circ f)(x) = f(f(x)) = f\left(\frac{4x}{x-1}\right)$$

Substitute $\frac{4x}{x-1}$ into the function $f(x) = \frac{4x}{x-1}$:

$$(f \circ f)(x) = \frac{4\left(\frac{4x}{x-1}\right)}{\left(\frac{4x}{x-1}\right)-1}$$

$$= \frac{\frac{16x}{x-1}}{\frac{4x - (x-1)}{x-1}}$$ $$= \frac{\frac{16x}{x-1}}{\frac{4x - x + 1}{x-1}}$$ $$= \frac{\frac{16x}{x-1}}{\frac{3x + 1}{x-1}}$$ $$= \frac{16x}{x-1} \cdot \frac{x-1}{3x+1}$$ $$(f \circ f)(x) = \boxed{\frac{16x}{3x+1}}$$

For $(f \circ f)(x)$ to be defined, two conditions must be met:

1. $x$ must be in the domain of $f$. This means $x \neq 1$.
2. $f(x)$ must be in the domain of $f$. This means $f(x) \neq 1$. $$\frac{4x}{x-1} \neq 1$$ $$4x \neq x-1$$ $$3x \neq -1$$ $$x \neq -\frac{1}{3}$$

Also, the denominator of the simplified $(f \circ f)(x)$ cannot be zero, so $3x+1 \neq 0 \implies x \neq -\frac{1}{3}$. Combining these conditions, the domain is $\boxed{x \in \mathbb{R}, x \neq 1, x \neq -\frac{1}{3}}$.

4. (ii) (a)
The statement $p \wedge q$ means "$p$ and $q$".
$p$: The principal goes to the ministry.
$q$: The principal will meet the minister.
Therefore, $p \wedge q$ in ordinary English is:
$\boxed{\text{The principal goes to the ministry and will meet the minister.}}$