Step-by-Step Solution

1. (i) Complete the figure:

   The figure should show points O, A, B, and C, with vectors OA = a, OB = b, and OC = 5a - 2b. It should also show vectors AC and BC.

2. (ii) Find vectors AC and BC:

   $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (5\vec{a} - 2\vec{b}) - \vec{a} = 4\vec{a} - 2\vec{b}$
   $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (5\vec{a} - 2\vec{b}) - \vec{b} = 5\vec{a} - 3\vec{b}$

3. (iii) (a) Find the angle between OA and OB and |a x b|:

   Given: $\vec{a} \cdot \vec{b} = 1$, $|\vec{a}| = 1$, $|\vec{b}| = 2$
   $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$
   $1 = (1)(2) \cos{\theta}$
   $\cos{\theta} = \frac{1}{2}$
   $\theta = \frac{\pi}{3}$ or 60 degrees.
   $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin{\theta} = (1)(2) \sin{\frac{\pi}{3}} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

4. (iii) (b) Find a unit vector perpendicular to (a+b) and (a-b):

   Given: $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$
   $\vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + (\hat{j} - \hat{k}) = 2\hat{i} + 3\hat{k}$
   $\vec{a} - \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{j} - \hat{k}) = 2\hat{i} - 2\hat{j} + 5\hat{k}$
   A vector perpendicular to both $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ is given by their cross product:
   $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 2 & -2 & 5 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(10 - 6) + \hat{k}(-4 - 0) = 6\hat{i} - 4\hat{j} - 4\hat{k}$
   Magnitude of the cross product: $|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{6^2 + (-4)^2 + (-4)^2} = \sqrt{36 + 16 + 16} = \sqrt{68} = 2\sqrt{17}$
   Unit vector: $\frac{6\hat{i} - 4\hat{j} - 4\hat{k}}{2\sqrt{17}} = \frac{3}{\sqrt{17}}\hat{i} - \frac{2}{\sqrt{17}}\hat{j} - \frac{2}{\sqrt{17}}\hat{k}$