If exp $\left\{\left({\tan }^{2}x-{\tan }^{4}x+{\tan }^{6}x-{\tan }^{8}x+\dots \right)\right.\left.{\log }_{e}16\right\},0<x<\pi /4,$ satisfies the quadratic equation $x^2-3x+2=0,$ then value of ${\cos }^{2}x+{\cos }^{4}x$ is

We have
${\tan }^{2}x-{\tan }^{4}x+{\tan }^{\circ }x-{\tan }^{\circ }x+\dots$
$=\frac{{\tan }^{2}x}{1-\left(-{\tan }^{2}x\right)}=\frac{{\tan }^{2}x}{{\sec }^{2}x}={\sin }^{2}x$
Therefore
$\begin{array}{l}y=\exp \left\{\left({\tan }^{2}x-{\tan }^{4}x+{\tan }^{6}x-{\tan }^{8}x+\dots \right){\log }_{e}16\right\}\\ =\exp \left\{\left({\sin }^{2}x\right){\log }_{e}16\right\}\\ =\exp \left\{{\log }_e\left({16}^{{\sin }^{2}x}\right)\right\}={16}^{{\sin }^{2}x}\end{array}$

As $y$ satisfies $x^2-3x+2=0,$ we get $y=1$or $y=2$
$\Rightarrow {16}^{{\sin }^{2}x}=1$ or ${16}^{{\sin }^{2}x}=2$
Since $0<x<\pi /4, 0<\sin x<1/\sqrt{2} \Rightarrow 0<{\sin }^{2}x<1/2$
$\therefore {16}^{{\sin }^{2}x}=1$ is not possible.
Thus,      ${16}^{{\sin }^{2}x}=2\Rightarrow {\sin }^{2}x=1/4$
Thus, ${\cos }^{2}x+{\cos }^{4}x=\left(1-{\sin }^{2}x\right)+{\left(1-{\sin }^{2}x\right)}^2=21/16$