1.6
distance up the incline (with kinetic friction present) ...
$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$
time up the incline ...
$t = \dfrac{v_f-v_0}{a}$
... where $v_f=0$ and $a = -g(\sin{\theta} + \mu \cos{\theta})$time down the incline ...
$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2\Delta x}{a}}$
note $v_0 = 0$, $\Delta x$ is the opposite of that found going up the incline and $a = -g(\sin{\theta} - \mu \cos{\theta})$