So, I know that since I have $F_y(t)$ I can say that: $$F_z(t) = P(Z\leq t) = P (8+2Y\leq t) = P (Y <= t-8/2) = F_y (t-8/2)$$
From this I have the distribution function so am I just differentiating $(t-8/2)^3$ then to get the pdf ?
Also I know the median is $P(Y\leq t)=0.5$ so previously I done that $t-8/2 = 0.5$ but again I am totally unsure and pretty sure it's wrong.
Any help would be great !
You are on the right track. You want to write the CDF as follows:
$ F_W(t) = F(W \leq t) = P(8+2Y \leq t) = P \Big(Y \leq \frac{t-8}{2} \Big) = F_Y \Big( \frac{t-8}{2} \Big) $
Next, you want to compute $ f_W(t) = \frac{dF_W(t)}{dt} = \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) $.
Case 1 $ (t<-1): $
Since $ F_Y(t) = 0 $ when $ t < -1 $, then $ \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = 0 $ when $ \frac{t-8}{2}<-1 \rightarrow t < 6 $. So $ f_W(t) = 0 $ when $ t < 6 $.
Case 2 $ (t>1): $
Since $ F_Y(t) = 1 $ when $ t > 1 $, then $ \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = 0 $ when $ \frac{t-8}{2}>1 \rightarrow t > 10 $. So $ f_W(t) = 0 $ when $ t < 10 $.
Case 3 $ (-1 \leq t \leq 1): $
Since $ F_Y(t) = \frac{t^3 + 1}{2} $ when $ (-1 \leq t \leq 1) $, then $ F_Y \Big(\frac{t-8}{2} \Big) = \frac{1}{2} \Big( \Big( \frac{t-8}{2} \Big)^3 + 1 \Big) $ when $ -1 \leq \frac{t-8}{2} \leq 1 \rightarrow 6 \leq t \leq 10 $. Finally, we want to differentiate with respect to t, i.e.
$ f_W(t) = \frac{dF_W(t)}{dt} = \frac{d}{dt} F_Y \Big( \frac{t-8}{2} \Big) = \frac{d}{dt} \Big( \frac{1}{2} \Big( \frac{t-8}{2} \Big)^3 + \frac{1}{2} \Big) = \frac{3}{2} \Big( \frac{t-8}{2} \Big)^2 \frac{1}{2} = \frac{3}{4} \Big( \frac{t-8}{2} \Big)^2 $
Median:
Since the median is the point $ t_0 $ where $ F_W(t_0) = \frac{1}{2} $ you can solve for it as follows:
$ F_W(t) = \frac{1}{2} \rightarrow F_W(t) = F_Y \Big( \frac{t-8}{2} \Big) = \frac{1}{2} \Big( \frac{t-8}{2} \Big)^3 + \frac{1}{2} = \frac{1}{2} \rightarrow t = 8 $.