This is a gamma distribution with r = 4 and
\begin{equation*}
F(x) = 1 - \sum_{k=0}^3 \frac{ (x/0.7)^k e^{-x/0.7}}{k!}
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
P(X \lt 3.5) \amp = F(3.5)\\
\amp = 1 - \frac{ (3.5/0.7)^0 e^{-3.5/0.7}}{0!} \\ - \frac{ (3.5/0.7)^1 e^{-3.5/0.7}}{1!} \\ - \frac{ (3.5/0.7)^2 e^{-3.5/0.7}}{2!} - \frac{ (3.5/0.7)^3 e^{-3.5/0.7}}{3!}\\
\amp = 1 - e^{-5} - 5 \cdot e^{-5} - \frac{25}{2} \cdot e^{-5} - \frac{125}{6} \cdot e^{-5}\\
\amp = 1 - e^{-5} \left ( 1 + 5 + 25/2 + 125/6 \right ) \approx 0.73497.
\end{aligned}
\end{equation*}