$$\begin{gathered} \mathrm{Statement}-\mathrm{I} : \mathrm{If} {\mathrm{x}}_{\mathrm{n}}=\cos \left(\mathrm{\pi }/2^{\mathrm{n}}\right)+\mathrm{i} \sin \left(\mathrm{\pi }/2^{\mathrm{n}}\right) \\ \mathrm{then} {\mathrm{x}}_1.{\mathrm{x}}_2.{\mathrm{x}}_3.......=-1 \\ \mathrm{Statement}-\mathrm{II} : {\mathrm{x}}_{\mathrm{n}}=\cos \left(\mathrm{\pi }/4^{\mathrm{n}}\right)+\mathrm{i} \sin \left(\mathrm{\pi }/4^{\mathrm{n}}\right) \\ \mathrm{then} {\mathrm{x}}_1.{\mathrm{x}}_2.{\mathrm{x}}_3.......\infty =-1 \end{gathered}$$

Statement I

$\begin{array}{rcl}{\mathrm{x}}_{\mathrm{n}}& =& \mathrm{cis} \frac{\mathrm{\pi }}{2^{\mathrm{n}}} \mathrm{then} {\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3.....=\mathrm{cis}\frac{\mathrm{\pi }}{2} \mathrm{cis}\frac{\mathrm{\pi }}{2^2} \mathrm{cis} \frac{\mathrm{\pi }}{2^3} ....\\ {\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3....& =& \mathrm{cis} \left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2^2}+\frac{\mathrm{\pi }}{2^3}+....\infty \right)\\ & =& \mathrm{cis}\frac{\mathrm{\pi }}{2}\left(\frac{1}{1-{\displaystyle \frac{1}{2}}}\right)=\mathrm{cis} \mathrm{\pi }=-1\end{array}$

Statement II

$\begin{array}{rcl}{\mathrm{x}}_{\mathrm{n}}& =& \mathrm{cis}\frac{\mathrm{\pi }}{4^{\mathrm{n}}}\\ {\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3.....\infty & =& \mathrm{cis}\frac{\mathrm{\pi }}{4} \mathrm{cis}\frac{\mathrm{\pi }}{4^2} \mathrm{cis}\frac{\mathrm{\pi }}{4^3} ....\infty \\ & =& \mathrm{cis} \frac{\mathrm{\pi }}{4}\left(1+\frac{1}{4}+\frac{1}{4^2}+....\infty \right)\\ & =& \mathrm{cis} \frac{\mathrm{\pi }}{4}\left(\frac{1}{1{\displaystyle \frac{-1}{4}}}\right)=\mathrm{cis} \frac{\mathrm{\pi }}{3}\end{array}$

$\therefore$ Only I is true.