1. From \(\sin(A-B)=1\) we know that \(A-B = 90^{\circ}\) because \(\sin\theta = 1\) only when \(\theta = 90^{\circ}+360^{\circ}k\). Since \(A>B\) the smallest positive value is taken: \(A-B = 90^{\circ}\).\
2. From \(2\cos(A+B)=1\) we get \(\cos(A+B)=\frac12\). In the interval \(0^{\circ}
3. Now we have a system of two linear equations:\
\[\begin{cases} A - B = 90^{\circ} \\ A + B = 60^{\circ} \end{cases}\]\
4. Add the two equations: \(2A = 150^{\circ}\) \(\Rightarrow\) \(A = 75^{\circ}\).\
5. Substitute \(A\) in \(A + B = 60^{\circ}\): \(75^{\circ}+B = 60^{\circ}\) \(\Rightarrow\) \(B = -15^{\circ}\).\
6. Verify the conditions: \(A>B\) (75° > -15°) and \(0^{\circ}Thus the required angles are \(A = 75^{\circ}\) and \(B = -15^{\circ}\).