I imagine this is a rather simple problem, but I'm having a bit of a hard time actually finding the answer.
$X \sim \mathrm{Exp}(0.2)$ and $W=g(X)$ given by
$g(X) = \begin{cases} X^{\frac{1}{3}} & \text{if }X \leq 8 \\ \frac{1}{7} (X-1)^2 -5 & \text{if }8 < X \leq 15 \\ 2X-7 & \text{if } 15 < X \end{cases}$
So if $x\in (0,2]$ then $P(W\leq x)=P(X\leq x^3)= \int_0 ^{x^3} (0.2) e^{-0.2 x} \, dx$. Then, $F_g(x) = \begin{cases} 0 & \text{if }x \leq 0 \\ \int_0 ^{x^{3}} (0.2) e^{-0.2 t} \, \text{d}t & \text{if }0 < x \leq 2 \\ \int_0 ^{\sqrt{7x+35} +1} (0.2) e^{-0.2 t} \, \text{d}t & \text{if } 2 < x \leq 23 \\ \int_0^{\frac{x + 7}{2}} (0.2) e^{-0.2 t} \, \text{d}t &\text{if } x > 23 \end{cases}$
Is this the correct approach? It seems reasonable to me that since $g(X)$ is "shifting" the probability of certain intervals in $X$ then the CDF is the same thing with the CDF of $X$, but I'm not sure if this is right.
For every increasing function $g:[0,+\infty)\to[0,+\infty)$ and every nonnegative $w$, $$[W\leqslant w]=[g(X)\leqslant w]=[X\leqslant g^{-1}(w)],$$ hence $$F_W=F_X\circ g^{-1}.$$ Now, your task is to check that the function $g$ you are considering is increasing.