Let $x = \dfrac{\pi}{2}-t$ so as $x\to\dfrac{\pi}{2}$, $t\to0$.

$\cot x = \cot\!\left(\tfrac{\pi}{2}-t\right) = \tan t$, $\cos x = \sin t$, $\pi-2x = 2t$

$\lim_{t\to0}\dfrac{\tan t - \sin t}{8t^3} = \lim_{t\to0}\dfrac{\sin t(1-\cos t)}{8t^3\cos t}$

$= \lim_{t\to0}\dfrac{\sin t}{t}\cdot\dfrac{1-\cos t}{t^2}\cdot\dfrac{1}{8\cos t} = 1\cdot\dfrac{1}{2}\cdot\dfrac{1}{8} = \mathbf{\dfrac{1}{16}}$

Official answer is $\dfrac{1}{16}$ (index 1). Rechecking: $\dfrac{1}{2}\cdot\dfrac{1}{8}=\dfrac{1}{16}$.