Less Talk, More Rock

$\sin(2x) = \frac {2}{5} \Rightarrow \sin(x) \cos(x) = \frac {1}{5}$

$\begin{aligned} \sin^8 (x) + \cos^8 (x) & = & \left ( \sin^2 (x) + \cos^2 (x) \right )^4 - 4 \sin^2 (x) \cos^2 (x) \left ( \sin^4 (x) + \cos^4 (x) \right ) - 6 \sin^4 (x) \cos^4 (x) \\ & = & 1 - 4 ( \sin (x) \cos (x) )^2 \cdot \left ( (\sin^2 (x) + \cos^2 (x) )^2 - 2\sin^2 (x) \cos^2 (x) \right ) - 6 ( \sin (x) \cos (x) )^4 \\ & = & 1 - 4 (\sin (x) \cos (x) )^2 \cdot \left ( 1 - 2 ( \sin (x) \cos (x) )^2 \right ) - 6 ( \sin (x) \cos (x) )^4 \\ & = & 1 - 4 \left ( \frac {1}{5} \right )^2 \cdot \left ( 1 - 2 \left ( \frac {1}{5} \right )^2 \right ) - 6 \left ( \frac {1}{5} \right )^4 \\ & = & \frac {527}{625} \\ \end{aligned}$

$\Rightarrow b - a = \boxed{98}$