The function $\text{f}$ is defined by $\text{f} : x \to \frac{1}{3}(e^{x-2}-1), x > 2$.

1. Set up an equation and make $x$ the subject to find the inverse function.

Let $y = \text{f}(x) = \frac{1}{3}(e^{x-2}-1)$.
$3y = e^{x-2}-1$
$3y + 1 = e^{x-2}$
Take the natural logarithm on both sides:
$\ln(3y+1) = x-2$
$x = 2 + \ln(3y+1)$

2. State the inverse function explicitly in terms of $x$.

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

3. Determine the domain of the inverse function by finding the range of the original function.

The domain of $\text{f}^{-1}$ is equal to the range of $\text{f}$.
Given the domain of $\text{f}$ is $x > 2$:
$x - 2 > 0$
$e^{x-2} > e^0 = 1$
$e^{x-2} - 1 > 0$
$\frac{1}{3}(e^{x-2} - 1) > 0$
Thus, the range of $\text{f}$ is $(0, \infty)$.
Therefore, the domain of $\text{f}^{-1}$ is $x > 0$.

1. State the geometric relationship that simplifies the equation.

Solutions to $\text{f}(x) = \text{f}^{-1}(x)$ occur where the graphs of $y = \text{f}(x)$ and $y = \text{f}^{-1}(x)$ intersect. Since these graphs are reflections of each other in the line $y = x$, their intersection points must lie on $y = x$.
Therefore, $\text{f}(x) = \text{f}^{-1}(x) \implies \text{f}(x) = x$.

2. Substitute the function into the simplified equation to derive the given result.

$\frac{1}{3}(e^{x-2}-1) = x$
$e^{x-2}-1 = 3x$
$e^{x-2} = 3x+1 \text{ (shown)}$

3. Solve the derived equation using a graphing calculator and select the valid root.

Using a graphing calculator to solve $e^{x-2} - 3x - 1 = 0$:
$x \approx -0.300 \text{ or } x \approx 4.718$
Since the domain of $\text{f}$ is $x > 2$, we reject $x \approx -0.300$.
$x = 4.72 \text{ (to 3 s.f.)}$

1. Determine the range of $\text{f}$ and compare it with the domain of $\text{g}$ to show $\text{gf}$ exists.

For $x > 2$, we have $x-2 > 0 \implies e^{x-2} > 1$.
Thus, $\text{f}(x) = \frac{1}{3}(e^{x-2}-1) > 0$.
The range of $\text{f}$ is $R_{\text{f}} = (0, \infty)$.
The domain of $\text{g}$ is $D_{\text{g}} = \mathbb{R} = (-\infty, \infty)$.
Since $R_{\text{f}} \subseteq D_{\text{g}}$, the composite function $\text{gf}$ exists.

2. Find the expression for the composite function $\text{gf}$.

$\text{gf}(x) = \text{g}(\text{f}(x))$
$= 1 + [\text{f}(x)]^2$
$= 1 + \left[ \frac{1}{3}(e^{x-2}-1) \right]^2$
$= 1 + \frac{1}{9}(e^{x-2}-1)^2$

3. State the domain and corresponding range of $\text{gf}$.

The domain of $\text{gf}$ is the same as the domain of $\text{f}$, so $D_{\text{gf}} = (2, \infty)$ or $x > 2$.
To find the range of $\text{gf}$, consider the values of $\text{g}(y)$ for $y \in R_{\text{f}} = (0, \infty)$.
For $y > 0$, $y^2 > 0 \implies 1+y^2 > 1$.
Thus, the range of $\text{gf}$ is $(1, \infty)$ or $\text{gf}(x) > 1$.