I may not be true. There is no enough information to prove or disprove $AF \sin \alpha = BE \sin \beta$.
II must be true. In $\Delta BCE$,
$\begin{array}{rcl}
\cos \alpha & = & \dfrac{BE}{CE} \\
BE & = & CE \cos \alpha
\end{array}$
In $\Delta ADF$,
$\begin{array}{rcl}
\cos \beta & = & \dfrac{AF}{DF} \\
AF & = & DF \cos \beta
\end{array}$
Since $E$ and $F$ divide $AB$ into three equals parts, then $AF = BE$. Hence, $CE \cos \alpha = DF \cos \beta$.
III must be true. In $\Delta BCE$,
$\begin{array}{rcl}
\tan \alpha & = & \dfrac{BC}{BE} \\
BE & = & \dfrac{BC}{\tan \alpha}
\end{array}$
In $\Delta ADF$,
$\begin{array}{rcl}
\tan \beta & = & \dfrac{AD}{AF} \\
AF & = & \dfrac{AD}{\tan \beta}
\end{array}$
Since $E$ and $F$ divide $AB$ into three equals parts, then $AF = BE$. Hence, we have
$\begin{array}{rcl}
\dfrac{BC}{\tan \alpha} & = & \dfrac{AD}{\tan \beta} \\
AD \tan \alpha & = & BC \tan \beta
\end{array}$
