A block is sent up a slope at 27° with a speed of 8.0 m/s…

Step 1. Going up. Both forces act down the slope: $a = 9.8\ \text{m/s}^2 \times (0.454 + 0.15 \times 0.891) = 5.76\ \text{m/s}^2$, a deceleration.

Step 2. Distance. $d = \dfrac{u^2}{2a} = \dfrac{64\ \text{m}^2/\text{s}^2}{2 \times 5.76\ \text{m/s}^2} = 5.56\ \text{m}$.

Step 3. Does it slide back? $\tan 27^\circ = 0.51$, more than $\mu_s = 0.30$. Static friction cannot hold it, so yes.

Step 4. Coming down. $a = 9.8\ \text{m/s}^2 \times (0.454 - 0.15 \times 0.891) = 3.14\ \text{m/s}^2$, so $v = \sqrt{2 \times 3.14\ \text{m/s}^2 \times 5.56\ \text{m}} = 5.91\ \text{m/s}$.