I must be true. The range of $A$
$\begin{array}{cl}
= & \delta – \alpha
\end{array}$
And the range of $B$
$\begin{array}{cl}
= & \mu – \alpha
\end{array}$
Since $\delta < \mu$, then we have
$\begin{array}{rcl}
\delta – \alpha & < & \mu - \alpha \\
\text{the range of }A & < & \text{the range of }B
\end{array}$
II must be true. Note that $\alpha < \beta < \gamma < \delta < \mu$, we have
$\begin{array}{cl}
& \text{the mean of }A-\text{ the mean of }B \\
= & \dfrac{\alpha+\beta+\gamma+\delta}{4} – \dfrac{\alpha + \cdots + \mu}{5} \\
= & \dfrac{5(\alpha + \cdots + \delta) – 4(\alpha + \cdots + \mu)}{20} \\
= & \dfrac{\alpha + \beta + \gamma + \delta – 4\mu}{20} \\
< & 0
\end{array}$
Therefore, the mean of $A<$ the mean of $B$.
III must be true. The median of $A$
$\begin{array}{cl}
= & \dfrac{\beta + \gamma}{2}
\end{array}$
The median of $B$
$\begin{array}{cl}
= & \gamma
\end{array}$
Since $\beta < \gamma$, we have
$\begin{array}{rcl}
\beta + \gamma & < & \gamma + \gamma \\
\beta + \gamma & < & 2 \gamma \\
\dfrac{\beta + \gamma}{2} & < & \gamma \\
\text{the median of }A & < & \text{the median of }B
\end{array}$
