If  $y={\left(0.64\right)}^{{\log }_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \infty )}$, then  $y=$

Let  $S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \infty$ is in G.P.

$=\frac{\frac{1}{3}}{1-\left(\frac{1}{3}\right)}(\because S_r=\frac{a}{1-r})$

$=\frac{1}{2}$

Now,  ${\log }_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots \cdots \infty )={\log }_{0.25}\left(\frac{1}{2}\right)={\log }_{\frac{25}{100}}\left(\frac{1}{2}\right)$

${\log }_{{\left(\frac{1}{2}\right)}^2}\frac{1}{2}=\frac{1}{2}{\log }_{\left(\frac{1}{2}\right)}\left(\frac{1}{2}\right)$

$=\frac{1}{2}\left(1\right)=\frac{1}{2}$

$\therefore y={\left(0.64\right)}^{{\log }_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots )}={\left(0.64\right)}^{\frac{1}{2}}$

$={\left({0.8}^2\right)}^{\frac{1}{2}}={\left(0.8\right)}^{2\times \frac{1}{2}}$

$=\left(0.8\right)$