Let $f:S\to\mathbb{R}$ be the function from the set of all right triangles into the set of real numbers, defined by $f(\Delta ABC)=\frac{h}{r}$, where $h$ is the height with respect to the hypotenuse and $r$ is the inscribed circle's radius. Find the image, $Im(f)$, of the function.
1. Given a right triangle $\Delta ABC$ with legs $a$ and $b$, and hypotenuse $c$, we need to find the image of the function $f(\Delta ABC) = \frac{h}{r}$, where $h$ is the height with respect to the hypotenuse and $r$ is the radius of the inscribed circle.

2. The height $h$ with respect to the hypotenuse $c$ can be expressed as:
   \[
   h = \frac{ab}{c}
   \]
   This follows from the area of the triangle, which can be written as $\frac{1}{2}ab$ and also as $\frac{1}{2}ch$. Equating these gives $h = \frac{ab}{c}$.

3. The radius $r$ of the inscribed circle can be expressed as:
   \[
   r = \frac{a + b - c}{2}
   \]
   This follows from the formula for the radius of the inscribed circle in a right triangle, which is given by:
   \[
   r = \frac{a + b - c}{2}
   \]

4. Therefore, the function $f(\Delta ABC)$ can be written as:
   \[
   f(\Delta ABC) = \frac{h}{r} = \frac{\frac{ab}{c}}{\frac{a + b - c}{2}} = \frac{2ab}{c(a + b - c)}
   \]

5. Simplifying further, we use the fact that $c = \sqrt{a^2 + b^2}$:
   \[
   f(\Delta ABC) = \frac{2ab}{\sqrt{a^2 + b^2}(a + b - \sqrt{a^2 + b^2})}
   \]

6. To simplify the expression, let $x = \frac{a}{b}$. Then $a = bx$ and $c = b\sqrt{x^2 + 1}$. Substituting these into the expression for $f(\Delta ABC)$, we get:
   \[
   f(\Delta ABC) = \frac{2b^2x}{b\sqrt{x^2 + 1}(b(x + 1) - b\sqrt{x^2 + 1})} = \frac{2x}{\sqrt{x^2 + 1}(x + 1 - \sqrt{x^2 + 1})}
   \]

7. Simplifying further, we get:
   \[
   f(\Delta ABC) = \frac{2x}{\sqrt{x^2 + 1}(x + 1 - \sqrt{x^2 + 1})} = \frac{2x}{(x + 1)\sqrt{x^2 + 1} - (x^2 + 1)}
   \]

8. Simplifying the denominator:
   \[
   (x + 1)\sqrt{x^2 + 1} - (x^2 + 1) = x\sqrt{x^2 + 1} + \sqrt{x^2 + 1} - x^2 - 1
   \]

9. Therefore, the function $g(x) = \frac{x + 1}{\sqrt{x^2 + 1}} + 1$ is symmetric, i.e., $g\left(\frac{1}{x}\right) = g(x)$ for all $x \in \mathbb{R}^+$.

10. We study $g(x)$ for $x \in [1, +\infty)$. The derivative of $g(x)$ is:
    \[
    g'(x) = \frac{1 - x}{(x^2 + 1)\sqrt{x^2 + 1}} \leq 0
    \]
    This shows that $g(x)$ is a non-increasing function.

11. Evaluating $g(x)$ at $x = 1$:
    \[
    g(1) = 1 + \frac{1 + 1}{\sqrt{1^2 + 1}} = 1 + \sqrt{2}
    \]

12. As $x \to +\infty$, $g(x) \to 2$. Therefore, $2 < g(x) \leq 1 + \sqrt{2}$ for all $x \in [1, +\infty)$.

13. Since $g(x)$ is continuous, every point in $(2, 1 + \sqrt{2}]$ is attained for some value of $g(x)$.

14. Hence, the image of the function $f$ is:
    \[
    Im(f) = (2, 1 + \sqrt{2}]
    \]

The final answer is $\boxed{(2, 1 + \sqrt{2}]}$