To determine the range of the function $$f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$$, let's start by simplifying the expression. Let $$\sin^2 \theta = x$$, so $$\cos^2 \theta = 1 - x$$. The function then transforms into:

$$ f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)} $$

Simplify the numerator and denominator separately:

Numerator: $$ x^2 + 3 - 3x $$

Denominator: $$ x^2 + 1 - x $$

Thus, the function becomes:

$$ f(x) = \frac{x^2 + 3 - 3x}{x^2 + 1 - x} = \frac{x^2 - 3x + 3}{x^2 - x + 1} $$

Next, we need to find the range of this function. Let's analyze the function by testing specific values of $$x$$ in the interval $$[0, 1]$$ (since $$\sin^2 \theta$$ ranges from 0 to 1):

When $$x = 0$$:

$$ f(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3 $$

When $$x = 1$$:

$$ f(1) = \frac{1^2 - 3(1) + 3}{1^2 - 1 + 1} = \frac{1 - 3 + 3}{1 - 1 + 1} = \frac{1}{1} = 1 $$

It appears that $$f(x)$$ achieves values within $$[1, 3]$$. To confirm this, we need to solve the quadratic inequality:

$$ 1 \leq \frac{x^2 - 3x + 3}{x^2 - x + 1} \leq 3 $$

By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:

$$ \alpha = 1 $$

$$ \beta = 3 $$

The common ratio of the infinite geometric progression is:

$$ \frac{\alpha}{\beta} = \frac{1}{3} $$

Given the first term $a = 64$, the sum $S$ of the infinite geometric progression can be given as:

$$ S = \frac{a}{1 - r} $$

Substituting the values $a = 64$ and $r = \frac{1}{3}$, we get:

$$ S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 $$

Therefore, the sum of the infinite geometric progression is 96.