Evaluate $\sec'' \frac{\pi}4 +\sec'' \frac{3\pi}4+\sec'' \frac{5\pi}4+\sec'' \frac{7\pi}4$. (Here $\sec''$ means the second derivative of $\sec$).
1. First, we need to find the second derivative of the secant function, $\sec x$. We start by finding the first derivative:
   \[
   \frac{d}{dx} \sec x = \sec x \tan x
   \]

2. Next, we find the second derivative by differentiating the first derivative:
   \[
   \frac{d^2}{dx^2} \sec x = \frac{d}{dx} (\sec x \tan x)
   \]
   Using the product rule, we get:
   \[
   \frac{d^2}{dx^2} \sec x = \frac{d}{dx} (\sec x) \cdot \tan x + \sec x \cdot \frac{d}{dx} (\tan x)
   \]
   \[
   \frac{d^2}{dx^2} \sec x = (\sec x \tan x) \cdot \tan x + \sec x \cdot \sec^2 x
   \]
   \[
   \frac{d^2}{dx^2} \sec x = \sec x \tan^2 x + \sec^3 x
   \]

3. Simplify the expression:
   \[
   \frac{d^2}{dx^2} \sec x = \sec x (\tan^2 x + \sec^2 x - 1)
   \]
   Using the identity $\sec^2 x = 1 + \tan^2 x$, we get:
   \[
   \frac{d^2}{dx^2} \sec x = \sec x (\tan^2 x + 1 + \tan^2 x - 1)
   \]
   \[
   \frac{d^2}{dx^2} \sec x = \sec x (2 \tan^2 x + 1)
   \]
   \[
   \frac{d^2}{dx^2} \sec x = 2 \sec^3 x - \sec x
   \]

4. Now, we need to evaluate this second derivative at the given points:
   \[
   \sec'' \left( \frac{\pi}{4} \right) + \sec'' \left( \frac{3\pi}{4} \right) + \sec'' \left( \frac{5\pi}{4} \right) + \sec'' \left( \frac{7\pi}{4} \right)
   \]

5. Evaluate $\sec x$ and $\tan x$ at each point:
   \[
   \sec \left( \frac{\pi}{4} \right) = \sqrt{2}, \quad \tan \left( \frac{\pi}{4} \right) = 1
   \]
   \[
   \sec \left( \frac{3\pi}{4} \right) = -\sqrt{2}, \quad \tan \left( \frac{3\pi}{4} \right) = -1
   \]
   \[
   \sec \left( \frac{5\pi}{4} \right) = -\sqrt{2}, \quad \tan \left( \frac{5\pi}{4} \right) = 1
   \]
   \[
   \sec \left( \frac{7\pi}{4} \right) = \sqrt{2}, \quad \tan \left( \frac{7\pi}{4} \right) = -1
   \]

6. Substitute these values into the second derivative formula:
   \[
   \sec'' \left( \frac{\pi}{4} \right) = 2 (\sqrt{2})^3 - \sqrt{2} = 2 \cdot 2\sqrt{2} - \sqrt{2} = 4\sqrt{2} - \sqrt{2} = 3\sqrt{2}
   \]
   \[
   \sec'' \left( \frac{3\pi}{4} \right) = 2 (-\sqrt{2})^3 - (-\sqrt{2}) = 2 \cdot (-2\sqrt{2}) + \sqrt{2} = -4\sqrt{2} + \sqrt{2} = -3\sqrt{2}
   \]
   \[
   \sec'' \left( \frac{5\pi}{4} \right) = 2 (-\sqrt{2})^3 - (-\sqrt{2}) = 2 \cdot (-2\sqrt{2}) + \sqrt{2} = -4\sqrt{2} + \sqrt{2} = -3\sqrt{2}
   \]
   \[
   \sec'' \left( \frac{7\pi}{4} \right) = 2 (\sqrt{2})^3 - \sqrt{2} = 2 \cdot 2\sqrt{2} - \sqrt{2} = 4\sqrt{2} - \sqrt{2} = 3\sqrt{2}
   \]

7. Add the results:
   \[
   3\sqrt{2} + (-3\sqrt{2}) + (-3\sqrt{2}) + 3\sqrt{2} = 0
   \]

The final answer is $\boxed{0}$.