$\begin{array}{rcl}
b & = & \sqrt{ac} \\
b^2 & = & ac
\end{array}$
I must be true.
$\begin{array}{rcl}
\log b^2 – \log a^2 & = & \log \dfrac{b^2}{a^2} \\
& = & \log \dfrac{ac}{a^2} \\
& = & \log \dfrac{c}{a}
\end{array}$
Also,
$\begin{array}{rcl}
\log c^2 – \log b^2 & = & \log \dfrac{c^2}{b^2} \\
& = & \log \dfrac{c^2}{ac} \\
& = & \log \dfrac{c}{a}
\end{array}$
Since $\log b^2 -\log a^2 = \log c^2 – \log b^2$, then $\log a^2$, $\log b^2$, $\log c^2$ is an arithmetic sequence.
II must be true.
$\begin{array}{rcl}
\dfrac{b^3}{a^3} & = & \left( \dfrac{b}{a} \right)^3 \\
& = & \left( \dfrac{\sqrt{ac}}{a} \right)^3 \\
& = & \left( \dfrac{\sqrt{c}}{\sqrt{a}} \right)^3
\end{array}$
Also,
$\begin{array}{rcl}
\dfrac{c^3}{a^3} & = & \left( \dfrac{c}{a} \right)^3 \\
& = & \left( \dfrac{c}{\sqrt{ac}} \right)^3 \\
& = & \left( \dfrac{\sqrt{c}}{\sqrt{a}} \right)^3
\end{array}$
Since $\dfrac{b^3}{a^3} = \dfrac{c^3}{a^3}$, then $a^3$, $b^3$, $c^3$ is a geometric sequence.
III may be false. Let $a=2$, $b=4$ and $c=8$. Note that $b=\sqrt{ac}$.
$\begin{array}{rcl}
\dfrac{4^b}{4^a} & = & \dfrac{4^4}{4^2} \\
& = & 4^{4-2} \\
& = & 4^2
\end{array}$
However,
$\begin{array}{rcl}
\dfrac{4^c}{4^b} & = & \dfrac{4^8}{4^4} \\
& = & 4^{8-4} \\
& = & 4^4
\end{array}$
Since $\dfrac{4^b}{4^a} \neq \dfrac{4^c}{4^b}$, then $4^a$, $4^b$, $4^c$ may not be a geometric sequence.
