Rationalise and simplify 3 1+ 3.√2

$$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$$

The simplified expression is $\boxed{6\sqrt{2}}$.

ii) $4\sqrt{18} - 2\sqrt{8}$

$$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$$ $$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$$

$$12\sqrt{2} - 4\sqrt{2} = 8\sqrt{2}$$

The simplified expression is $\boxed{8\sqrt{2}}$.

b) Given that $\sqrt{a} + \sqrt{b} = 5$ and $\sqrt{a} - \sqrt{b} = 1$, find the values of $a$ and $b$.

Let $x = \sqrt{a}$ and $y = \sqrt{b}$.

$$x + y = 5 \quad \text{(Equation 1)}$$ $$x - y = 1 \quad \text{(Equation 2)}$$

$$(x + y) + (x - y) = 5 + 1$$ $$2x = 6$$ $$x = 3$$

$$3 + y = 5$$ $$y = 5 - 3$$ $$y = 2$$

Since $x = \sqrt{a}$, we have $\sqrt{a} = 3 \implies a = 3^2 = 9$.
Since $y = \sqrt{b}$, we have $\sqrt{b} = 2 \implies b = 2^2 = 4$.
The values are $\boxed{a=9, b=4}$.

QUESTION 2 (10 marks)
Theme 1 & 2: Operations and Rationalisation

a) Calculate $3\sqrt{2} \times 2\sqrt{8}$ and express your answer in simplest form.

$$3\sqrt{2} \times 2\sqrt{8} = (3 \times 2) \times (\sqrt{2} \times \sqrt{8})$$ $$= 6 \times \sqrt{2 \times 8}$$ $$= 6 \times \sqrt{16}$$

$$= 6 \times 4$$ $$= 24$$

The simplified product is $\boxed{24}$.

b) Rationalise the denominator of $\frac{5}{\sqrt{5}}$ and simplify.

$$= \frac{5\sqrt{5}}{5}$$ $$= \sqrt{5}$$

The rationalised and simplified expression is $\boxed{\sqrt{5}}$.

c) Simplify and rationalise $\frac{2+\sqrt{3}}{\sqrt{3}}$. Give your answer in the form $a + b\sqrt{3}$ where $a$ and $b$ are rational numbers.

$$= \frac{(2\sqrt{3}) + (\sqrt{3} \times \sqrt{3})}{3}$$ $$= \frac{2\sqrt{3} + 3}{3}$$

$$= \frac{3}{3} + \frac{2\sqrt{3}}{3}$$ $$= 1 + \frac{2}{3}\sqrt{3}$$

The simplified and rationalised expression is $\boxed{1 + \frac{2}{3}\sqrt{3}}$.

QUESTION 3 (10 marks)
Theme 2: Rationalisation with Binomial Denominators

a) Rationalise the denominator of $\frac{1}{3-\sqrt{5}}$ and simplify.

Use the difference of squares formula $(x-y)(x+y) = x^2 - y^2$ for the denominator.

$$= \frac{1 \times (3+\sqrt{5})}{3^2 - (\sqrt{5})^2}$$ $$= \frac{3+\sqrt{5}}{9 - 5}$$ $$= \frac{3+\sqrt{5}}{4}$$

The rationalised and simplified expression is $\boxed{\frac{3+\sqrt{5}}{4}}$.

b) Show that $\frac{\sqrt{48}+2\sqrt{27}}{\sqrt{12}} = 8$.

$$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$ $$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$$ $$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$$

$$= \frac{4\sqrt{3} + 6\sqrt{3}}{2\sqrt{3}}$$ $$= \frac{10\sqrt{3}}{2\sqrt{3}}$$

$$= \frac{10}{2}$$ $$= 5$$

The left-hand side simplifies to 5. Since $5 \neq 8$, the statement $\frac{\sqrt{48}+2\sqrt{27}}{\sqrt{12}} = 8$ is incorrect. The expression simplifies to $\boxed{5}$.