Let $X$ be a random variable with probability distribution function:
$F(x) = \begin{cases} 1 - ae^{-x/5} : x \ge 0 \\ 0 : x < 0 \end{cases} $
Let $ Z = F(X)$, find the probability distribution function of $Z$.
My attempt: $F(z) = P(Z<z) = P(F(X) <z) = P(X <F^{-1}(z)) = F(F^{-1}(z)) = \begin{cases} z : z \ge 0 \\ 0 : z < 0 \end{cases} $
Is my answer correct and rigorous enough?
Regardless of the continuous distribution of the random variable $X$, the distribution of $F(X)$ will always be the same. This is known as the probability integral transform, also called the inverse transform sampling.
In your final step, the correct conclusion should be
$$P(F(X)\le z)=\begin{cases}0 &,\text{ if }z\le0\\z&,\text{ if }0<z<1\\1&,\text{ if }z\ge1\end{cases}$$
So we see that $F(X)$ has the Uniform distribution over $(0,1)$.
You should denote the distribution function of $Z$ by something like $F_Z(\cdot)$ since you already use $F$ to denote the CDF of $X$. One thing you could also add while deriving the distribution function of $F(X)$ is that the inverse of $F$ exists as $F$ is strictly increasing and continuous.