You have done most of the analysis, so I will be to a great extent repeating what you know. We want to find an expression for $\Pr(V\le w)$.
In general, $V\le w$ iff $|X-\lambda| \le w$ iff $X-\lambda\le w$ and $X-\lambda\ge -w$, that is, iff
$$\lambda-w \le X\le \lambda+w.$$
There are three cases to consider, (i) $w\le 0$; (ii) $0\lt w\le \lambda$; and (ii) $w \gt \lambda$.
Case (i): This is trivial: if $w\le 0$ then $\Pr(V\le w)=0$.
Case (ii): We want $\Pr(X\le \lambda+w)-\Pr(X\lt \lambda -w)$. This is
$$(1-e^{-\lambda(\lambda+w)})-(1-e^{-\lambda(\lambda-w)}).$$
There is some immediate simplification, to $e^{-\lambda(\lambda-w)}-e^{-\lambda(\lambda+w)}$, and there are various alternate ways to rewrite things, by introducing the hyperbolic sine.
Case (iii): This one is easier. We simply want $\Pr(X\le \lambda+w)$. For $w\ge -lambda$, this is
$$1-e^{-\lambda(\lambda+w)}.$$
We could have set up the calculations using integrals, but since we already know that $F_X(x)=1-e^{-\lambda x}$ (when $x\gt 0$) there is no need to do that.
Now that we have the cdf of $V$, it is straightforward to find the density. For $w\le 0$, we have $f_V(w)=0$. For $0\lt w\lt \lambda$, we have $f_V(w)=\lambda e^{-\lambda(\lambda-w)}+\lambda e^{-\lambda(\lambda+w)}$. Finally, for $w\gt \lambda$ we have $f_V(w)=\lambda e^{-\lambda(\lambda+w)}$.
Remark: Suppose that we did not have a nice expression for the cdf of $X$. That happens, for example, with the normal, and a number of other distributions. We could still find the density function by setting up our probabilities as integrals, and differentiating under the integral sign.
Best Answer
The four necessary properties of a CDF are:
You have a suspected-to-be cumulative distribution function: $$F(x) = \begin{cases} 0 & : x \leq 0 \\ x^2/3 & : 0\leq x\leq 3 \\ 1& : 3 \leq x\end{cases}$$
Although you've only provided the middle piece. However the other pieces are actually important. With them included, are you satisfied that the four properties hold ?