Let $$\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \infty$$ and $$\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \infty$$. Then the value of$$(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)}$$ is equal to :

The infinite series $$\alpha = \frac14 + \frac18 + \frac1{16} + \cdots$$ is a geometric progression with first term $$a = \frac14$$ and common ratio $$r = \frac12$$.

The sum to infinity of a GP is $$S_\infty = \frac{a}{1-r}$$. Hence
$$\alpha = \frac{\tfrac14}{1-\tfrac12} = \frac{\tfrac14}{\tfrac12} = \frac12$$.

Similarly, $$\beta = \frac13 + \frac19 + \frac1{27} + \cdots$$ has $$a = \frac13, r = \frac13$$, so
$$\beta = \frac{\tfrac13}{1-\tfrac13} = \frac{\tfrac13}{\tfrac23} = \frac12$$.

The required expression is
$$\left(0.2\right)^{\log_{\sqrt5}(\alpha)} + \left(0.04\right)^{\log_5(\beta)}.$$

Substituting $$\alpha = \beta = \frac12$$ gives
$$\left(0.2\right)^{\log_{\sqrt5}\!\left(\tfrac12\right)} + \left(0.04\right)^{\log_5\!\left(\tfrac12\right)}.$$(1)

Convert the numbers to powers of 5:
$$0.2 = \frac15 = 5^{-1}, \qquad 0.04 = \frac1{25} = 5^{-2}, \qquad \sqrt5 = 5^{\tfrac12}.$$

First exponent:
$$\log_{\sqrt5}\!\left(\tfrac12\right) = \frac{\ln(\tfrac12)}{\ln(5^{\tfrac12})} = \frac{\ln(\tfrac12)}{\tfrac12\ln5} = 2\,\frac{\ln(\tfrac12)}{\ln5} = 2\log_5\!\left(\tfrac12\right).$$

Let $$x = \log_5\!\left(\tfrac12\right).$$ Then $$x = -\log_5 2.$$

Rewrite each term in (1):
$$\left(5^{-1}\right)^{2x} + \left(5^{-2}\right)^{x}.$$

Using $$(a^m)^n = a^{mn}$$,
$$\left(5^{-1}\right)^{2x} = 5^{-2x}, \qquad \left(5^{-2}\right)^{x} = 5^{-2x}.$$(2)

Because both terms in (2) are identical, evaluate one of them:
$$-2x = -2\!\left(-\log_5 2\right) = 2\log_5 2.$$

Thus
$$5^{-2x} = 5^{2\log_5 2}.$$

Using the identity $$b^{\log_b k} = k$$ and $$2\log_5 2 = \log_5 2^2 = \log_5 4$$,
$$5^{2\log_5 2} = 5^{\log_5 4} = 4.$$(3)

Each of the two terms in (1) equals $$4$$, so the sum is
$$4 + 4 = 8.$$

Therefore, the value of the given expression is $$8$$.

Option C which is: $$8$$