Let $$k$$ and $$K$$ be the minimum and the maximum values of the function $$f(x) = \frac{(1+x)^{0.6}}{1+x^{0.6}}$$ in $$[0, 1]$$, respectively, then the ordered pair $$(k, K)$$ is equal to:

We are given the function $$ f(x) = \frac{(1+x)^{0.6}}{1+x^{0.6}} $$ defined on the interval $$[0, 1]$$. We need to find the minimum value $$ k $$ and the maximum value $$ K $$ of this function in $$[0, 1]$$, and then the ordered pair $$(k, K)$$.

First, evaluate the function at the endpoints of the interval.

At $$ x = 0 $$:
$$ f(0) = \frac{(1+0)^{0.6}}{1+0^{0.6}} = \frac{1^{0.6}}{1+0} = \frac{1}{1} = 1 $$.

At $$ x = 1 $$:
$$ f(1) = \frac{(1+1)^{0.6}}{1+1^{0.6}} = \frac{2^{0.6}}{1+1} = \frac{2^{0.6}}{2} = 2^{0.6} \cdot 2^{-1} = 2^{0.6 - 1} = 2^{-0.4} $$.

So, $$ f(0) = 1 $$ and $$ f(1) = 2^{-0.4} $$.

Next, determine if there are any critical points in $$(0, 1)$$ by finding the derivative of $$ f(x) $$ and setting it to zero. The function is $$ f(x) = \frac{(1+x)^{0.6}}{1+x^{0.6}} $$. Use the quotient rule: let $$ u = (1+x)^{0.6} $$ and $$ v = 1 + x^{0.6} $$, so $$ f(x) = \frac{u}{v} $$. Then, $$ f'(x) = \frac{u' v - u v'}{v^2} $$.

Compute the derivatives:
$$ u' = 0.6 (1+x)^{-0.4} $$,
$$ v' = 0.6 x^{-0.4} $$.

Substitute into the quotient rule:
$$ f'(x) = \frac{ [0.6 (1+x)^{-0.4}] \cdot [1 + x^{0.6}] - [(1+x)^{0.6}] \cdot [0.6 x^{-0.4}] }{ (1 + x^{0.6})^2 } $$.

Factor out 0.6:
$$ f'(x) = \frac{0.6}{ (1 + x^{0.6})^2 } \left[ (1+x)^{-0.4} (1 + x^{0.6}) - (1+x)^{0.6} x^{-0.4} \right] $$.

Simplify the expression inside the brackets. Let $$ g(x) = (1+x)^{-0.4} (1 + x^{0.6}) - (1+x)^{0.6} x^{-0.4} $$. This can be written as:
$$ g(x) = (1+x)^{-0.4} + (1+x)^{-0.4} x^{0.6} - (1+x)^{0.6} x^{-0.4} $$.

Consider the last two terms:
$$ (1+x)^{-0.4} x^{0.6} - (1+x)^{0.6} x^{-0.4} = (1+x)^{-0.4} x^{-0.4} \left[ x^{0.6} x^{0.4} - (1+x)^{0.6} (1+x)^{0.4} \right] = (1+x)^{-0.4} x^{-0.4} \left[ x^{1} - (1+x)^{1} \right] = (1+x)^{-0.4} x^{-0.4} (x - 1 - x) = (1+x)^{-0.4} x^{-0.4} (-1) = - (1+x)^{-0.4} x^{-0.4} $$.

So, $$ g(x) = (1+x)^{-0.4} + \left( - (1+x)^{-0.4} x^{-0.4} \right) = (1+x)^{-0.4} \left( 1 - x^{-0.4} \right) $$.

Now, $$ 1 - x^{-0.4} = 1 - \frac{1}{x^{0.4}} = -\frac{1 - x^{0.4}}{x^{0.4}} $$. Thus,
$$ g(x) = (1+x)^{-0.4} \left( -\frac{1 - x^{0.4}}{x^{0.4}} \right) $$.

Substitute back into $$ f'(x) $$:
$$ f'(x) = \frac{0.6}{ (1 + x^{0.6})^2 } \cdot (1+x)^{-0.4} \left( -\frac{1 - x^{0.4}}{x^{0.4}} \right) = -\frac{0.6}{ (1 + x^{0.6})^2 } \cdot (1+x)^{-0.4} \cdot \frac{1 - x^{0.4}}{x^{0.4}} $$.

For $$ x \in (0, 1) $$, all factors in the derivative are positive except the negative sign and $$ 1 - x^{0.4} $$:
- 0.6 is positive.
- $$ (1 + x^{0.6})^2 $$ is always positive.
- $$ (1+x)^{-0.4} $$ is positive.
- $$ x^{0.4} $$ is positive.
- Since $$ x < 1 $$, $$ x^{0.4} < 1 $$, so $$ 1 - x^{0.4} > 0 $$.

Thus, the entire expression for $$ f'(x) $$ is negative for all $$ x \in (0, 1) $$. Therefore, $$ f(x) $$ is strictly decreasing on $$[0, 1]$$.

For a strictly decreasing function on $$[0, 1]$$, the maximum value occurs at the left endpoint $$ x = 0 $$ and the minimum value occurs at the right endpoint $$ x = 1 $$.

So,
Maximum $$ K = f(0) = 1 $$,
Minimum $$ k = f(1) = 2^{-0.4} $$.

The ordered pair is $$ (k, K) = (2^{-0.4}, 1) $$.

Comparing with the options:
A. $$ (2^{-0.4}, 1) $$
B. $$ (2^{-0.6}, 1) $$
C. $$ (2^{-0.4}, 2^{0.6}) $$
D. $$ (1, 2^{0.6}) $$

The pair $$ (2^{-0.4}, 1) $$ matches option A.

Hence, the correct answer is Option A.