Let $$f : R \to R$$ and $$g : R \to R$$ be two functions defined by $$f(x) = \log_e(x^2 + 1) - e^{-x} + 1$$ and $$g(x) = \frac{1 - 2e^{2x}}{e^x}$$. Then, for which of the following range of $$\alpha$$, the inequality $$f\left(g\left(\frac{\left(\alpha-1\right)^2}{3}\right)\right)>f\left(g\left(\alpha-\frac{5}{3}\right)\right)$$ holds?

Given,

$$f(x)=\log_e(x^2+1)-e^{-x}+1$$

and

$$g(x)=\frac{1-2e^{2x}}{e^x}=e^{-x}-2e^x$$

We need to solve

$$f\left(g\left(\frac{(\alpha-1)^2}{3}\right)\right)>f\left(g\left(\alpha-\frac53\right)\right)$$

First, find the nature of $$f$$.

Differentiating,

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

Now,

$$\frac{2x}{x^2+1}\ge -1$$

and

$$e^{-x}>0$$

Hence,

$$f'(x)>0\qquad \forall x\in\mathbb R$$

Therefore, $$f(x)$$ is strictly increasing.

So,

$$f(A)>f(B)\iff A>B$$

Hence, the given inequality becomes

$$g\left(\frac{(\alpha-1)^2}{3}\right)>g\left(\alpha-\frac53\right)$$

Now,

$$g(x)=e^{-x}-2e^x$$

Differentiating,

$$g'(x)=-e^{-x}-2e^x<0$$

Therefore, $$g(x)$$ is strictly decreasing.

Hence,

$$g(u)>g(v)\iff u<v$$

So,

$$\frac{(\alpha-1)^2}{3}<\alpha-\frac53$$

Multiplying by $$3,$$

$$ (\alpha-1)^2<3\alpha-5 $$

$$ \alpha^2-2\alpha+1<3\alpha-5 $$

$$ \alpha^2-5\alpha+6<0 $$

$$ (\alpha-2)(\alpha-3)<0 $$

Therefore,

$$2<\alpha<3$$

Hence, the required range of $$\alpha$$ is $$\boxed{(2,3)}$$.