Evaluate (27/64)^(1/3) and Factorise completely pq - 3p - q2 + 3q. and Simplify 8x + 4 - 3(2x - 3).

A power of $\frac{1}{3}$ is equivalent to taking the cube root.

$$\left(\frac{27}{64}\right)^{\frac{1}{3}} = \frac{27^{\frac{1}{3}}}{64^{\frac{1}{3}}} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}}$$

$$\sqrt[3]{27} = 3 \quad \text{because} \quad 3 \times 3 \times 3 = 27$$ $$\sqrt[3]{64} = 4 \quad \text{because} \quad 4 \times 4 \times 4 = 64$$

$$\frac{\sqrt[3]{27}}{\sqrt[3]{64}} = \frac{3}{4}$$

The evaluated expression is:

$$\boxed{\frac{3}{4}}$$

2. Factorise completely $pq - 3p - q^2 + 3q$.

From the first group, factor out $p$:

$$p(q - 3)$$

From the second group, factor out $-q$:

$$-q(q - 3)$$

The expression now is:

$$p(q - 3) - q(q - 3)$$

$$(q - 3)(p - q)$$

The completely factorised expression is:

$$\boxed{(q - 3)(p - q)}$$

3. Simplify $8x + 4 - 3(2x - 3)$.

$$8x + 4 - 3(2x) - 3(-3)$$ $$8x + 4 - 6x + 9$$

Combine the $x$ terms:

$$8x - 6x = 2x$$

Combine the constant terms:

$$4 + 9 = 13$$

$$2x + 13$$

The simplified expression is:

$$\boxed{2x + 13}$$