From the given information, it is apparent that $a_{n+1}\in (-\frac{\pi}{2}, \frac{\pi}{2})$ holds for every $n\in Z^+$, and \begin{equation} \tan{a_{n+1}} = \sec{a_n} \end{equation} Because $\sec{a_1} > 0$, we can conclude $a_{n+1} \in (0, \frac{\pi}{2})$ for every $n$. From the last relation, we have $\tan^2{a_{n+1}} = \sec^2{a_{n}}= 1 + \tan^2{a_n}$. Or $$\tan^2{a_n} = (n-1)+\tan^2{a_1} = (n-1)+\frac{1}{3} \implies \tan{a_n}=\frac{3n-2}{3}$$ Therefore: \begin{align} &\sin{a_1}\cdot\sin{a_2}\cdots\sin{a_m}\\ =&\frac{\tan{a_1}}{\sec{a_1}}\cdot\frac{\tan{a_2}}{\sec{a_2}}\cdots\frac{\tan{a_m}}{\sec{a_m}}\\ =&\frac{\tan{a_1}}{\tan{a_2}}\cdot\frac{\tan{a_2}}{\tan{a_3}}\cdots\frac{\tan{a_m}}{\tan{a_{m+1}}}\\ =&\frac{\tan{a_1}}{\tan{a_{m+1}}}\\ =&\sqrt{\frac{1}{3m+1}} \end{align} Setting $$\sqrt{\frac{1}{3m+1}}=\frac{1}{100}$$ yields $m=\boxed{3333}$.