For the exponential distribution random variable $X$, the pdf is
\[
f(x;\lambda)=\begin{cases}
\lambda e^{-\lambda x}, & x\geqslant 0,\\
0, &\text{otherwise}
\end{cases}
\]
where $\lambda > 0$.
The cdf of $X$ is
\[
F(x;\lambda)=\begin{cases}
0, & x < 0,\\
1-e^{-\lambda x}, & x\geqslant 0.
\end{cases}
\]
(a)
\[
P(X\leqslant 100)=F(100;\lambda)=1-e^{-\lambda\cdot 100}
\]
Here $\lambda=.01386$.
Thus the probability of $X$ being at most $200$ is
\[
P(X\leqslant 100)=F(100;\lambda)=1-e^{-0.01386\cdot 100}=1-e^{-1.386}\approx 0.74992640
\]
The probability that the distance $X$ is at most $200$ is
\[
P(X\leqslant 200)=F(200;\lambda)=1-e^{-0.01386\cdot 200}=1-e^{-2.772}\approx 0.93746319
\]
The probability that the distance $X$ is between $100$ and $200$ is
\[
P(100\leqslant X\leqslant 200)=F(200;\lambda)-F(100;\lambda)\approx 0.93746319-0.74992640=0.18753679
\]
(b)
The mean distance is equal to
\[E(X)=\mu=\alpha\beta=1\cdot\frac{1}{\lambda}=\frac{1}{\lambda}=\frac{1}{0.01386}\approx 72.15007215\]
The variance of $X$ is
\[
V(X)=\sigma^2=\alpha\beta^2=1\cdot(\frac{1}{\lambda})^2=\frac{1}{\lambda^2}\approx 5205.63291126
\]
So, the mean and standard deviation of exponential distribution equal $\frac{1}{\lambda}\approx 72.15007215$.
The probability that the distance exceeds the mean distance by more than $2$ standard deviations is
\[
\begin{split}
P(X-\mu\geqslant 2\sigma)&=P(X-72.15007215\geqslant 2\times 72.15007215)=P(X\geqslant 3\times 72.15007215)=P(X\geqslant 216.45021645)\\
&=1-P(X\leqslant 216.45021645)=1-(1-e^{-\lambda\cdot 216.45021645})\\
&=e^{-0.01386\times 216.45021645}=e^{-2.9999999999970}\\
&\approx 0.04978707
\end{split}
\]
(c)
Suppose the median distance is $x_0$, then $F(x_0;\lambda)=\frac{1}{2}$. That is,
\[
\begin{split}
&1-e^{-\lambda x_0}=\frac{1}{2}\\
\Rightarrow&e^{-\lambda x_0}=\frac{1}{2}\\
\Rightarrow&-\lambda x_0=\ln(\frac{1}{2})\\
\Rightarrow&\lambda x_0=\ln 2\\
\Rightarrow& x_0=\frac{\ln 2}{\lambda}\approx\frac{0.69314717}{0.01386}=50.01061833
\end{split}
\]
