It is given that $\cos A = \frac{1}{2}$ and $\sin B = -\frac{1}{\sqrt{2}}$ where $0^\circ < A < 90^\circ$ and $180^\circ < B < 270^\circ$. Find, without using a calculator, the exact value of $\cos(A - B)$, leaving your answer in the form $p\sqrt{2} + q\sqrt{6}$, where $p$ and $q$ are real numbers.

1. Find the exact value of $\sin A$.

Since $0^\circ < A < 90^\circ$, $\sin A > 0$.
$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

2. Find the exact value of $\cos B$.

Since $180^\circ < B < 270^\circ$, $\cos B < 0$.
$\cos B = -\sqrt{1 - \sin^2 B} = -\sqrt{1 - \left(-\frac{1}{\sqrt{2}}\right)^2} = -\sqrt{1 - \frac{1}{2}} = -\frac{1}{\sqrt{2}}$

3. Apply the addition formula for $\cos(A - B)$.

$\cos(A - B) = \cos A \cos B + \sin A \sin B$
$\cos(A - B) = \left(\frac{1}{2}\right)\left(-\frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{\sqrt{2}}\right)$
$= -\frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}}$
$= \frac{-1 - \sqrt{3}}{2\sqrt{2}}$

4. Rationalise the denominator to express the answer in the required form.

Multiply numerator and denominator by $\sqrt{2}$:
$\cos(A - B) = \frac{(-1 - \sqrt{3})\sqrt{2}}{2\sqrt{2}\sqrt{2}}$
$= \frac{-\sqrt{2} - \sqrt{6}}{4}$
$= -\frac{1}{4}\sqrt{2} - \frac{1}{4}\sqrt{6}$