Consider the situation such that $ n$ circles with radius $ r$ are contained between 2 concentric circles with radius $ 1,\ 1\minus{}2r$ without intersecting one anoyher.Let $ r$ vary with being fixed $ n\geq 2.$

(1) Show that $ r$ should be satisfied $ 0<r\leq \frac{\sin \frac{\pi}{n}}{1\plus{}\sin \frac{\pi}{n}}.$

(2) Find the values of $ r$ for which the sum of the areas of $ n\plus{}2$ circles is minimized.
### Part 1: Show that \( 0 < r \leq \frac{\sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}} \)

1. **Cyclic \( n \)-gon and radius relation**:
   - The centers of the \( n \) small circles form a regular \( n \)-gon inscribed in a circle of radius \( 1 - r \).
   - The side length \( 2a \) of this \( n \)-gon is given by \( 2a = 2(1 - r) \sin \frac{\pi}{n} \).

2. **Non-intersecting condition**:
   - For the small circles not to intersect, the distance between any two centers must be at least \( 2r \).
   - Therefore, \( 2(1 - r) \sin \frac{\pi}{n} \geq 2r \).

3. **Solving the inequality**:
   \[
   (1 - r) \sin \frac{\pi}{n} \geq r
   \]
   \[
   \sin \frac{\pi}{n} - r \sin \frac{\pi}{n} \geq r
   \]
   \[
   \sin \frac{\pi}{n} \geq r (1 + \sin \frac{\pi}{n})
   \]
   \[
   r \leq \frac{\sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}}
   \]

4. **Conclusion**:
   - Since \( r \) must be positive, we have \( 0 < r \leq \frac{\sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}} \).

### Part 2: Find the values of \( r \) for which the sum of the areas of \( n + 2 \) circles is minimized

1. **Sum of areas**:
   - The total area \( A \) of the \( n \) small circles and the two larger circles is:
     \[
     A = n \pi r^2 + \pi (1^2) + \pi ((1 - 2r)^2)
     \]
   - Simplifying the expression:
     \[
     A = n \pi r^2 + \pi + \pi (1 - 4r + 4r^2)
     \]
     \[
     A = n \pi r^2 + \pi + \pi - 4 \pi r + 4 \pi r^2
     \]
     \[
     A = (n + 4) \pi r^2 - 4 \pi r + 2 \pi
     \]

2. **Minimizing the quadratic function**:
   - The quadratic function \( f(r) = (n + 4) r^2 - 4 r + 2 \) is minimized at:
     \[
     r = \frac{-b}{2a} = \frac{4}{2(n + 4)} = \frac{2}{n + 4}
     \]

3. **Verification of \( r \) within bounds**:
   - We need to check if \( r = \frac{2}{n + 4} \) satisfies \( 0 < r \leq \frac{\sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}} \).
   - We need to show:
     \[
     \frac{2}{n + 4} \leq \frac{\sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}}
     \]
   - This simplifies to:
     \[
     2 (1 + \sin \frac{\pi}{n}) \leq (n + 4) \sin \frac{\pi}{n}
     \]
     \[
     2 + 2 \sin \frac{\pi}{n} \leq n \sin \frac{\pi}{n} + 4 \sin \frac{\pi}{n}
     \]
     \[
     2 \leq n \sin \frac{\pi}{n}
     \]
   - Since \( \sin \frac{\pi}{n} \) is positive and increasing for \( n \geq 2 \), and \( \sin \frac{\pi}{2} = 1 \), the inequality holds for \( n \geq 2 \).

Conclusion:
\[
\boxed{r = \frac{2}{n + 4}}
\]