Simplify surds, solve quadratic equations, evaluate logarithms, and find the gradient of a function.

$$
\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}
$$
$$
\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}
$$

$$
6\sqrt{2} - 4\sqrt{2} = (6-4)\sqrt{2} = 2\sqrt{2}
$$
The simplified expression is $\boxed{2\sqrt{2}}$.

$$
x_1 = \frac{3 + 5}{4} = \frac{8}{4} = 2
$$
$$
x_2 = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}
$$
The solutions are $\boxed{x = 2, x = -\frac{1}{2}}$.

$$
\log_{10}(1000) = 3 \quad (\text{since } 10^3 = 1000)
$$
$$
\log_{10}(10) = 1 \quad (\text{since } 10^1 = 10)
$$

$$
3 - 1 = 2
$$
The evaluated expression is $\boxed{2}$.

$$
\frac{dy}{dx} = \frac{d}{dx}(3x^2 - 2x) = 6x - 2
$$

$$
\frac{dy}{dx}\Big|_{x=1} = 6(1) - 2 = 6 - 2 = 4
$$
The gradient at $x=1$ is $\boxed{4}$.

$$
\text{det} = (3)(4) - (2)(1)
$$
$$
\text{det} = 12 - 2
$$
$$
\text{det} = 10
$$
The determinant of the matrix is $\boxed{10}$.

$$
\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ
$$

In Quadrant I: $x_1 = 60^\circ$  
In Quadrant II: $x_2 = 180^\circ - 60^\circ = 120^\circ$  
The solutions are $\boxed{x = 60^\circ, 120^\circ}$.

$$
x \cdot x + x \cdot (-3) + 2 \cdot x + 2 \cdot (-3)
$$
$$
x^2 - 3x + 2x - 6
$$

$$
x^2 - x - 6
$$
The expanded expression is $\boxed{x^2 - x - 6}$.

$$
\text{Sum} = 2 + 4 + 6 + 8 = 20
$$

$$
\text{Mean} = \frac{20}{4} = 5
$$
The mean is $\boxed{5}$.

$$
\int 2x^1 \,dx = 2 \cdot \frac{x^{1+1}}{1+1} + C
$$
$$
= 2 \cdot \frac{x^2}{2} + C
$$
$$
= x^2 + C
$$
The integral is $\boxed{x^2 + C}$.

$$
x + 2 = 0 \implies x = -2
$$
$$
x + 3 = 0 \implies x = -3
$$
The solutions are $\boxed{x = -2, x = -3}$.

**b)** Hence find the sum and product of the roots (3 marks)

$$
\text{Sum} = (-2) + (-3) = -5
$$
Alternatively, using the formula for sum of roots $(-\frac{b}{a})$:
$$
\text{Sum} = -\frac{5}{1} = -5
$$

$$
\text{Product} = (-2) \times (-3) = 6
$$
Alternatively, using the formula for product of roots $(\frac{c}{a})$:
$$
\text{Product} = \frac{6}{1} = 6
$$
The sum of the roots is $\boxed{-5}$ and the product of the roots is $\boxed{6}$.

$$
3 \times 2 = 6
$$

$$
x^2 \cdot x^1 = x^{2+1} = x^3
$$

$$
y^1 \cdot y^3 = y^{1+3} = y^4
$$

$$
6x^3y^4
$$
The simplified expression is $\boxed{6x^3y^4}$.

**b)** Express your answer in index form (2 marks)  
The answer from part (a) is already in index form: $\boxed{6x^3y^4}$.

$$
y = 5 - x \quad (3)
$$

$$
x^2 + (5 - x)^2 = 13
$$
$$
x^2 + (25 - 10x + x^2) = 13
$$
$$
2x^2 - 10x + 25 = 13
$$

$$
2x^2 - 10x + 12 = 0
$$

$$
x^2 - 5x + 6 = 0
$$

$$
(x - 2)(x - 3) = 0
$$

$$
x - 2 = 0 \implies x = 2
$$
$$
x - 3 = 0 \implies x = 3
$$

If $x = 2$: $y = 5 - 2 = 3$  
If $x = 3$: $y = 5 - 3 = 2$  
The solutions are $\boxed{x=2, y=3}$ and $\boxed{x=3, y=2}$.

**b)** Find $xy$ (2 marks)

If $x=2, y=3$:
$$
xy = 2 \times 3 = 6
$$
If $x=3, y=2$:
$$
xy = 3 \times 2 = 6
$$
The value of $xy$ is $\boxed{6}$.

$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2
$$

**b)** Find the midpoint of the line (2 marks)

$$
M = \left(\frac{2 + 4}{2}, \frac{3 + 7}{2}\right)
$$
$$
M = \left(\frac{6}{2}, \frac{10}{2}\right)
$$
$$
M = (3, 5)
$$
The midpoint of the line is $\boxed{(3, 5)}$.