In other words, gives two independent random variables $X$ and $Y$, distributed according to hypoexponential distribution with parameters $\{ w_1, w_2, \ldots, w_n \}$ and $\{v_1,v_2, \ldots, v_m \}$ respectively, you are asking to determine the distribution of $Z=X-Y$.
Let $\Theta_X$, and $\Theta_Y$ denote matrices from the probability density functions of respective hypoexponential distributions, see wiki page:
$$
f_X(x) = - \langle\vec{\alpha}_n, \exp(x \Theta_X) \Theta_X, \vec{1}_n \rangle \cdot [ x > 0 ] \qquad \qquad
f_Y(y) = - \langle\vec{\alpha}_m, \exp(y \Theta_Y) \Theta_Y, \vec{1}_m \rangle \cdot [ y > 0 ]
$$
where $(\alpha_n)_i = \delta_{i,1}$, $ (\vec{1}_n)_i = 1$ and $(\alpha_m)_j = \delta_{j,1}$, $ (\vec{1}_m)_j = 1$, $i=1,\ldots,n$, and $j=1,\ldots,m$.
Then
$$ \begin{eqnarray}
f_Z(z) &=& \int_{-\infty}^\infty f_X(z+y) f_Y(y) \mathrm{d} y =
\int_{\max(-z,0)}^\infty f_X(z+y) f_Y(y) \mathrm{d} y \\
&=&
\int_{\max(-z,0)}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(z+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{y \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\
&=&
\int_{0}^\infty \langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{(\max(z,0)+y) \Theta_X} \Theta_X \right) \otimes \left( \mathrm{e}^{(\max(-z,0) + y) \Theta_Y} \Theta_Y\right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle \mathrm{d} y \\
&=&
\langle \vec{\alpha}_n \otimes \vec{\alpha}_m, \left( \mathrm{e}^{\max(z,0) \Theta_X} \otimes \mathrm{e}^{(\max(-z,0) ) \Theta_Y} \right) \cdot \left( \int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y \right), \vec{\mathbf{1}}_n \otimes \vec{\mathbf{1}}_m \rangle
\end{eqnarray}
$$
The formula above tells the density function for $Z$ will be piecewise, much like Laplace distribution, with functional form of $X$ variate for $z>0$ and functional form of $Y$ variate for $z<0$.
Example:
Consider an example with $n=2$ and $m=2$, and $\{w_1,w_2\} = \{1,2\}$, and $\{v_1,v_2\} = \{1,1\}$. Corresponding matrices are
$$
\Theta_X = \left( \begin{array}{cc} -1 & 1 \\ 0 & -2 \end{array} \right) \qquad
\Theta_Y = \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right)
$$
Then
$$
\exp\left(x \Theta_X \right) = \left(
\begin{array}{cc}
e^{-x} & e^{-x}-e^{-2 x} \\
0 & e^{-2 x} \\
\end{array}
\right) \qquad
\exp\left(y \Theta_Y \right) = \left(
\begin{array}{cc}
e^{-y} & e^{-y} y \\
0 & e^{-y} \\
\end{array}
\right)
$$
$$
\exp\left(x \Theta_X \right) \Theta_X =
\left(
\begin{array}{cc}
-e^{-x} & 2 e^{-2 x}-e^{-x} \\
0 & -2 e^{-2 x} \\
\end{array}
\right)
\qquad
\exp\left(y \Theta_Y \right) \Theta_Y =
\left(
\begin{array}{cc}
-e^{-y} & e^{-y}-e^{-y} y \\
0 & -e^{-y} \\
\end{array}
\right)
$$
Using Kronecker product,
$$
\int_{0}^\infty \mathrm{e}^{y \Theta_X} \Theta_X \otimes \mathrm{e}^{y \Theta_Y} \Theta_Y \mathrm{d} y = \left(
\begin{array}{cccc}
\frac{1}{2} & -\frac{1}{4} & -\frac{1}{6} & \frac{7}{36} \\
0 & \frac{1}{2} & 0 & -\frac{1}{6} \\
0 & 0 & \frac{2}{3} & -\frac{4}{9} \\
0 & 0 & 0 & \frac{2}{3} \\
\end{array}
\right)
$$
Combining things, with little algebra we get:
$$
f_Z(z) = \left\{
\begin{array}{cc}
\frac{5}{18} & z=0 \\
\frac{1}{18} \mathrm{e}^{-z} \left(9-4 \mathrm{e}^{-z}\right) & z>0 \\
\frac{1}{18} \mathrm{e}^z (5-6 z) & z < 0 \\
\end{array} \right.
$$
a) Some calculus needed. For the probability that $X\gt 5$, you need to find $\int_5^\infty \frac{1}{5}e^{-x/5}\,dx$.
For the mean, probably you are expected to use the fact that if $f_X(x)$ is the density function, then the mean is $\int_{-\infty}^\infty xf_X(x)\,dx$. In our case that comes down to evaluating $\int_0^\infty \frac{x}{5}e^{-x/5}\,dx$. The antiderivative you need is usually calculated by integration by parts.
b) he cumulative distribution function $F_X(x)$ of $X$ is $0$ if $x\lt 0$. For $x\ge 0$, it is given by
$$F_X(x)=\int_0^x \frac{1}{5}e^{-t/5}\,dt.$$
Now that we have $F_X$, note that $\Pr(X\gt 7)=1-\Pr(X\le 7)=1-F_X(7)$.
To go on, unfortunately I must reveal that for $x\ge 0$, it turns out that $F_X(x)=1-e^{-x/5}$.
To find the median $m$, we need to solve the equation
$$1-e^{-m/5}=\frac{1}{2}.$$
In solving this, you will need to use the (natural) logarithm function.
c) The $25$-th percentile is the point $a$ such that $\Pr(X\le a)=0.25$. To find $a$ you proceed much like in the calculation of the median.
The $75$-th percentile $b$ is computed in a similar way. Finally, calculate $b-a$.
Best Answer
You can directly compute the mean absolute deviation from the median in the following way $$ \text{Mean absolute deviation from the median}=\int|x-0|\frac{1}{2}e^{-\left|x\right|}\, dx=2\int_{x\geq 0}x\frac{1}{2}e^{-x}\, dx \\=\Gamma{(1)}=1. $$
So why your answer is different?
Median absolute deviation is a different quantity than the mean absolute deviation from the median. Hope this helps.