Find all pairs $(x,y)$ of real numbers that satisfy the system \begin{align*} x \cdot \sqrt{1-y^2} &=\frac14 \left(\sqrt3+1 \right),
\\ y \cdot \sqrt{1-x^2} &= \frac14 \left( \sqrt3 -1 \right).
\end{align*}
1. Given the system of equations:
   \[
   x \cdot \sqrt{1-y^2} = \frac{1}{4} \left(\sqrt{3} + 1 \right),
   \]
   \[
   y \cdot \sqrt{1-x^2} = \frac{1}{4} \left( \sqrt{3} - 1 \right).
   \]

2. Notice that both \(x\) and \(y\) are positive and \(x, y < 1\). Therefore, there exist real numbers \(\alpha\) and \(\beta\) such that \(x = \cos \alpha\) and \(y = \cos \beta\).

3. Substituting \(x = \cos \alpha\) and \(y = \cos \beta\) into the system, we get:
   \[
   \cos \alpha \cdot \sin \beta = \frac{1}{4} (\sqrt{3} + 1),
   \]
   \[
   \cos \beta \cdot \sin \alpha = \frac{1}{4} (\sqrt{3} - 1).
   \]

4. Using the trigonometric identity \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\), we can add the two equations:
   \[
   \cos \alpha \sin \beta + \cos \beta \sin \alpha = \frac{1}{4} (\sqrt{3} + 1) + \frac{1}{4} (\sqrt{3} - 1).
   \]

5. Simplifying the right-hand side:
   \[
   \sin(\alpha + \beta) = \frac{1}{4} (\sqrt{3} + 1 + \sqrt{3} - 1) = \frac{1}{4} (2\sqrt{3}) = \frac{\sqrt{3}}{2}.
   \]

6. The value \(\sin(\alpha + \beta) = \frac{\sqrt{3}}{2}\) corresponds to \(\alpha + \beta = 60^\circ\) or \(\alpha + \beta = 120^\circ\).

7. Similarly, using the trigonometric identity \(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha\), we can subtract the two equations:
   \[
   \cos \alpha \sin \beta - \cos \beta \sin \alpha = \frac{1}{4} (\sqrt{3} + 1) - \frac{1}{4} (\sqrt{3} - 1).
   \]

8. Simplifying the right-hand side:
   \[
   \sin(\beta - \alpha) = \frac{1}{4} (\sqrt{3} + 1 - \sqrt{3} + 1) = \frac{1}{4} (2) = \frac{1}{2}.
   \]

9. The value \(\sin(\beta - \alpha) = \frac{1}{2}\) corresponds to \(\beta - \alpha = 30^\circ\) or \(\beta - \alpha = 150^\circ\).

10. Solving the system of equations \(\alpha + \beta = 60^\circ\) and \(\beta - \alpha = 30^\circ\):
    \[
    \alpha + \beta = 60^\circ,
    \]
    \[
    \beta - \alpha = 30^\circ.
    \]

11. Adding these two equations:
    \[
    2\beta = 90^\circ \implies \beta = 45^\circ.
    \]

12. Substituting \(\beta = 45^\circ\) into \(\alpha + \beta = 60^\circ\):
    \[
    \alpha + 45^\circ = 60^\circ \implies \alpha = 15^\circ.
    \]

13. Therefore, \(x = \cos 15^\circ\) and \(y = \cos 45^\circ\):
    \[
    x = \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},
    \]
    \[
    y = \cos 45^\circ = \frac{\sqrt{2}}{2}.
    \]

14. Similarly, solving the system of equations \(\alpha + \beta = 120^\circ\) and \(\beta - \alpha = 30^\circ\):
    \[
    \alpha + \beta = 120^\circ,
    \]
    \[
    \beta - \alpha = 30^\circ.
    \]

15. Adding these two equations:
    \[
    2\beta = 150^\circ \implies \beta = 75^\circ.
    \]

16. Substituting \(\beta = 75^\circ\) into \(\alpha + \beta = 120^\circ\):
    \[
    \alpha + 75^\circ = 120^\circ \implies \alpha = 45^\circ.
    \]

17. Therefore, \(x = \cos 45^\circ\) and \(y = \cos 75^\circ\):
    \[
    x = \cos 45^\circ = \frac{\sqrt{2}}{2},
    \]
    \[
    y = \cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}.
    \]

Conclusion:
The pairs \((x, y)\) that satisfy the system are:
\[
\left( \frac{\sqrt{6} + \sqrt{2}}{4}, \frac{\sqrt{2}}{2} \right) \quad \text{and} \quad \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{6} - \sqrt{2}}{4} \right).
\]

The final answer is \( \boxed{ \left( \frac{\sqrt{6} + \sqrt{2}}{4}, \frac{\sqrt{2}}{2} \right) } \) and \(\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{6} - \sqrt{2}}{4} \right)\).