If point $M(x,y)$ lies on the line with equation $y=x+2$ and $1<y<3$, calculate the value of 
$A=\sqrt{y^2-8x}+\sqrt{y^2+2x+5}$
1. Given the point \( M(x, y) \) lies on the line with equation \( y = x + 2 \) and \( 1 < y < 3 \), we can substitute \( y = x + 2 \) into the inequality \( 1 < y < 3 \):
   \[
   1 < x + 2 < 3
   \]
   Subtracting 2 from all parts of the inequality:
   \[
   -1 < x < 1
   \]

2. We need to calculate the value of \( A = \sqrt{y^2 - 8x} + \sqrt{y^2 + 2x + 5} \). Since \( y = x + 2 \), we substitute \( y \) in the expression for \( A \):
   \[
   A = \sqrt{(x + 2)^2 - 8x} + \sqrt{(x + 2)^2 + 2x + 5}
   \]

3. Simplify each term inside the square roots:
   \[
   (x + 2)^2 = x^2 + 4x + 4
   \]
   Therefore:
   \[
   \sqrt{(x + 2)^2 - 8x} = \sqrt{x^2 + 4x + 4 - 8x} = \sqrt{x^2 - 4x + 4} = \sqrt{(x - 2)^2} = |x - 2|
   \]
   Since \( -1 < x < 1 \), \( x - 2 \) is always negative, so \( |x - 2| = 2 - x \).

4. For the second term:
   \[
   \sqrt{(x + 2)^2 + 2x + 5} = \sqrt{x^2 + 4x + 4 + 2x + 5} = \sqrt{x^2 + 6x + 9} = \sqrt{(x + 3)^2} = |x + 3|
   \]
   Since \( -1 < x < 1 \), \( x + 3 \) is always positive, so \( |x + 3| = x + 3 \).

5. Combine the simplified terms:
   \[
   A = |x - 2| + |x + 3| = (2 - x) + (x + 3) = 2 - x + x + 3 = 5
   \]

The final answer is \( \boxed{5} \).