Let $a$ and $r$ be the first term and the common ratio of the geometric sequence. Then we have
$\begin{array}{rcl}
ar^5 & = & 216 \ \ldots \unicode{x2460} \\
ar^7 & = & 96 \ \ldots \unicode{x2461}
\end{array}$
$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
r^2 & = & \dfrac{4}{9} \\
r & = & \pm \dfrac{2}{3}
\end{array}$
For $r=\dfrac{2}{3}$,
$\begin{array}{rcl}
a (\dfrac{2}{3})^5 & = & 216 \\
a & = & \dfrac{6561}{4}
\end{array}$
For $r=\dfrac{-2}{3}$,
$\begin{array}{rcl}
a (\dfrac{-2}{3})^5 & = & 216 \\
a & = & \dfrac{-6561}{4}
\end{array}$
I may not be true. If $a = \dfrac{-6561}{4}$ and $r =\dfrac{-2}{3}$, we have
$\begin{array}{rcl}
x_3 & = & \dfrac{-6561}{4} \times \left( \dfrac{-2}{3} \right)^2 \\
& = & -729
\end{array}$
II must be true. For $r= \pm\dfrac{2}{3}$, we have
$\begin{array}{rcl}
\dfrac{x_5}{x_7} & = & \dfrac{ar^4}{ar^6} \\
& = & \dfrac{1}{r^2} \\
& = & \dfrac{1}{(\pm \frac{2}{3})^2} \\
& = & \dfrac{9}{4} \\
& > & 1
\end{array}$
III must be true. Note that the new first term and the new common ratio are $\dfrac{2187}{2}$ and $\dfrac{4}{9}$ respectively. Hence, we have
$\begin{array}{rcl}
x_2 + x_4 + \cdots + x_{2n} & = & \dfrac{\frac{2187}{2}}{1-\frac{4}{9}} \\
& = & \dfrac{19683}{10} \\
& < & 2015
\end{array}$
