First thing to note is that there was a syllabus change in 2003 - whilst doing old questions still is very useful, make sure that you don't do them in preference to the more recent ones!

You should find that the stationary points satisfy $a \cos bx = b \sin bx$, so $\tan bx = \frac a b$ and $bx = \tan^{-1} (\frac a b) + n \pi$.

To make things a little tidier, let $\tan^{-1} (\frac a b)=k $. The values of the function at the turning points are then:
$\text{f}_n(x) = \text{e}^{\frac a b (k+n \pi)}\cos(k+n \pi)$

To show that they form a GP you can consider the ratio $\frac {\text{f}_{n+1}(x)} {\text{f}_{n}(x)}$. This gives (after cancelling bits):
\[
\frac {\text{f}_{n+1}(x)} {\text{f}_{n}(x)} = \frac {\text{e}^{\frac a b \pi} \cos (k + (n+1)\pi)}{\cos(k+n\pi)}
\]

Using the compound angle formula for $\cos(A+B)$ and noting that $\sin(n\pi)=0$ and $\cos(n \pi) = (-1)^n$ will give the required result.