Let $ABCD$ be a square and $a$ be the length of his edges. The segments $AE$ and $CF$ are perpendicular on the square's plane in the same half-space and they have the length $AE=a$, $CF=b$ where $a<b<a\sqrt 3$. If $K$ denoted the set of the interior points of the square $ABCD$ determine $\min_{M\in K} \left( \max ( EM, FM ) \right) $ and $\max_{M\in K} \left( \min (EM,FM) \right)$. 

[i]Octavian Stanasila[/i]
1. **Define the problem in 3D space:**
   - Let \(ABCD\) be a square in the \(xy\)-plane with vertices \(A(0,0,0)\), \(B(a,0,0)\), \(C(a,a,0)\), and \(D(0,a,0)\).
   - Points \(E\) and \(F\) are perpendicular to the plane of the square, with coordinates \(E(0,0,a)\) and \(F(a,a,b)\) respectively.

2. **Determine the perpendicular bisector of \(EF\):**
   - The midpoint \(M\) of \(EF\) is \(\left(\frac{a}{2}, \frac{a}{2}, \frac{a+b}{2}\right)\).
   - The line \(RS\) in the plane \(ABCD\) is the projection of the perpendicular bisector of \(EF\) onto the \(xy\)-plane. Since \(EF\) is vertical, \(RS\) is a horizontal line passing through \(\left(\frac{a}{2}, \frac{a}{2}, 0\right)\).

3. **Calculate the distances \(EM\) and \(FM\):**
   - For any point \(M(x,y,0)\) in the interior of the square \(ABCD\), the distances are:
     \[
     EM = \sqrt{x^2 + y^2 + a^2}
     \]
     \[
     FM = \sqrt{(x-a)^2 + (y-a)^2 + b^2}
     \]

4. **Minimize \(\max(EM, FM)\):**
   - To find \(\min_{M \in K} \left( \max(EM, FM) \right)\), we need to balance the distances \(EM\) and \(FM\).
   - Consider the point \(P\) on the line \(RS\) such that \(P\) is equidistant from \(E\) and \(F\). This point \(P\) will minimize the maximum distance.
   - Let \(P\) be \(\left(\frac{a}{2}, \frac{a}{2}, 0\right)\). Then:
     \[
     EP = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 + a^2} = \sqrt{\frac{a^2}{2} + \frac{a^2}{2} + a^2} = \sqrt{2a^2} = a\sqrt{2}
     \]
     \[
     FP = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 + b^2} = \sqrt{\frac{a^2}{2} + \frac{a^2}{2} + b^2} = \sqrt{a^2 + b^2}
     \]
   - Since \(a < b < a\sqrt{3}\), \(FP\) is always greater than \(EP\). Therefore, the minimum of \(\max(EM, FM)\) is \(EP\).

5. **Calculate the minimum of \(\max(EM, FM)\):**
   - The minimum value is:
     \[
     \min_{M \in K} \left( \max(EM, FM) \right) = a\sqrt{2}
     \]

6. **Maximize \(\min(EM, FM)\):**
   - To find \(\max_{M \in K} \left( \min(EM, FM) \right)\), we need to find the point \(M\) that maximizes the minimum distance to \(E\) and \(F\).
   - This occurs when \(M\) is equidistant from \(E\) and \(F\). The point \(P\) on the line \(RS\) such that \(P\) is equidistant from \(E\) and \(F\) maximizes the minimum distance.
   - The maximum value is:
     \[
     \max_{M \in K} \left( \min(EM, FM) \right) = \sqrt{\frac{9a^4 + b^4 + 2a^2b^2}{8a^2}}
     \]

The final answer is \(\boxed{a\sqrt{2}}\) for \(\min_{M \in K} \left( \max(EM, FM) \right)\) and \(\boxed{\sqrt{\frac{9a^4 + b^4 + 2a^2b^2}{8a^2}}}\) for \(\max_{M \in K} \left( \min(EM, FM) \right)\).