Let α=(1)/(4)+(1)/(8)+(1)/(16)+ ∞ and β=(1)/(3)+(1)/(9)+(1)/(27)+ ∞...

The sum of an infinite geometric progression is given by $S_{\infty} = \dfrac{a}{1-r}$.

For $\alpha$, $a = \dfrac{1}{4}$ and $r = \dfrac{1}{2}$:
$\alpha = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}} = \dfrac{1}{2}$

For $\beta$, $a = \dfrac{1}{3}$ and $r = \dfrac{1}{3}$:
$\beta = \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = \dfrac{1}{2}$

Now, evaluating the first term:
$(0.2)^{\log_{\sqrt{5}}(\alpha)} = \left(5^{-1}\right)^{\log_{5^{1/2}}(1/2)}$
Using the property $\log_{a^k}(x) = \dfrac{1}{k}\log_a(x)$:
$\log_{5^{1/2}}(1/2) = 2\log_5(1/2) = \log_5(1/4)$
Thus, \left(5^{-1}\right)^{\log_5(1/4)} = 5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4$<br><br>Evaluating the second term:<br>$(0.04)^{\log_5(\beta)} = \left(5^{-2}\right)^{\log_5(1/2)}$<br>\left(5^{-2}\right)^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4$

Adding both terms:
$(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)} = 4 + 4 = 8$

Answer: $8$