The functions $\text{f}$ and $\text{g}$ are defined by
$\text{f} : x \mapsto \frac{1}{2}\text{e}^{1-x^2}, \ x \in \mathbb{R}, \ x \le 1 \text{ and}$
$\text{g} : x \mapsto \sqrt{1 - \ln x}, \ x \in \mathbb{R}, \ 0 < x \le \text{e}.$

1. Find the range of $\text{f}$.

For $x \le 1$, the maximum value of the exponent $1 - x^2$ occurs at $x = 0$, giving a maximum of $1$. As $x \to -\infty$, $1 - x^2 \to -\infty$.
Thus, $1 - x^2 \le 1$.
Since $\text{f}(x) = \frac{1}{2}\text{e}^{1-x^2}$ and the exponential function is strictly positive, we have $0 < \text{f}(x) \le \frac{1}{2}\text{e}^1$.
Therefore, the range of $\text{f}$ is $\text{R}_\text{f} = (0, \frac{1}{2}\text{e}]$.

2. Verify the condition for the existence of $\text{gf}$.

The domain of $\text{g}$ is given as $\text{D}_\text{g} = (0, \text{e}]$.
Since $\frac{1}{2}\text{e} < \text{e}$, it follows that $(0, \frac{1}{2}\text{e}] \subseteq (0, \text{e}]$.
Therefore, $\text{R}_\text{f} \subseteq \text{D}_\text{g}$, which proves that the composite function $\text{gf}$ exists.

3. Find the range of $\text{gf}$ by mapping $\text{R}_\text{f}$ through $\text{g}$.

The range of $\text{gf}$ is the image of $\text{R}_\text{f}$ under $\text{g}$. We evaluate $\text{g}(x) = \sqrt{1 - \ln x}$ for $x \in (0, \frac{1}{2}\text{e}]$.
At the upper bound, $x = \frac{1}{2}\text{e}$:
$\text{g}\left(\frac{1}{2}\text{e}\right) = \sqrt{1 - \ln\left(\frac{1}{2}\text{e}\right)} = \sqrt{1 - (\ln \text{e} - \ln 2)} = \sqrt{\ln 2}$
As $x \to 0^+$, $\ln x \to -\infty$, so $1 - \ln x \to \infty$, giving $\text{g}(x) \to \infty$.
Since $y = \ln x$ is increasing, $\text{g}(x)$ is a strictly decreasing function for $x > 0$. Thus, the minimum value is $\sqrt{\ln 2}$ and the function increases without bound.
The range of $\text{gf}$ is $[\sqrt{\ln 2}, \infty)$.

1. Evaluate the function at two distinct points in the domain to test for a one-to-one relationship.

Consider $x = 1$ and $x = -1$, both of which are within the domain of $\text{f}$ ($x \le 1$).

$\text{f}(1) = \frac{1}{2}\text{e}^{1-(1)^2} = \frac{1}{2}\text{e}^0 = \frac{1}{2}$

$\text{f}(-1) = \frac{1}{2}\text{e}^{1-(-1)^2} = \frac{1}{2}\text{e}^0 = \frac{1}{2}$

2. Conclude whether the inverse function exists based on the definition of a one-to-one function.

Since $\text{f}(1) = \text{f}(-1)$ but $1 \neq -1$, the function $\text{f}$ is not one-to-one (it is many-to-one).

Therefore, the inverse function $\text{f}^{-1}$ does not exist.

1. Determine the largest value of $b$ for which the function is one-to-one.

The function $\text{f}(x) = \frac{1}{2}\text{e}^{1-x^2}$ has a stationary point where the exponent $1-x^2$ is at its maximum.
The maximum of $1-x^2$ occurs at $x = 0$.
For $\text{f}(x)$ to be a one-to-one function (which is required for an inverse to exist) on the domain $(-\infty, b]$, the interval cannot extend past the turning point at $x = 0$.
Therefore, the largest value of $b$ is $0$.

2. Find the algebraic rule for the inverse function.

Let $y = \frac{1}{2}\text{e}^{1-x^2}$.
$2y = \text{e}^{1-x^2}$
$\ln(2y) = 1 - x^2$
$x^2 = 1 - \ln(2y)$
$x = \pm\sqrt{1 - \ln(2y)}$
Since the restricted domain is $x \le 0$, we reject the positive root:
$x = -\sqrt{1 - \ln(2y)}$
Therefore, the rule for the inverse function is $\text{f}^{-1}(x) = -\sqrt{1 - \ln(2x)}$.

3. Determine the domain of the inverse function by finding the range of the restricted function $\text{f}$.

The domain of $\text{f}^{-1}$ corresponds to the range of $\text{f}$ on its restricted domain $(-\infty, 0]$.
For $x \le 0$, $\text{f}(x)$ is strictly increasing.
When $x = 0$, $\text{f}(0) = \frac{1}{2}\text{e}^1 = \frac{\text{e}}{2}$.
As $x \to -\infty$, $x^2 \to \infty$, so $\text{e}^{1-x^2} \to 0$, which means $\text{f}(x) \to 0$.
Thus, the range of $\text{f}$ is $\left(0, \frac{\text{e}}{2}\right]$.
The domain of $\text{f}^{-1}$ is $0 < x \le \frac{\text{e}}{2}$.

4. Define the inverse function clearly.

Combining the rule and the domain, the formal definition is:
$\text{f}^{-1} : x \mapsto -\sqrt{1 - \ln(2x)}, \ x \in \mathbb{R}, \ 0 < x \le \frac{\text{e}}{2}$

1. Apply the transformations to the equation of the function.

Scaling parallel to the $x$-axis by a factor of 2 replaces $x$ with $\frac{x}{2}$. Reflecting in the $x$-axis multiplies the entire function by $-1$.

$\text{h}(x) = -\text{g}\left(\frac{x}{2}\right) = -\sqrt{1 - \ln\left(\frac{x}{2}\right)}$

2. Determine the new domain after the transformation.

The original domain for $\text{g}$ is $0 < x \le \text{e}$.
A scaling parallel to the $x$-axis by a factor of 2 multiplies all the original $x$-values by 2.

$0 < \frac{x}{2} \le \text{e} \implies 0 < x \le 2\text{e}$

3. Define $\text{h}$ in the required similar form.

Combining the transformed rule and domain in formal mapping notation:

$\text{h} : x \mapsto -\sqrt{1 - \ln\left(\frac{x}{2}\right)}, \ x \in \mathbb{R}, \ 0 < x \le 2\text{e}$