If $A+B=60°$ then ${\cos }^{2}A+\cos B^2-\cos A\cos B=$

$\Rightarrow \cos \left[A+B\right]=\cos 60$

$\Rightarrow \cos A.\cos B-\sin A.\sin B=\frac{1}{2}$

$\Rightarrow 2\cos A.\cos B-1=2\sin A.\sin B$

Square on both sides

$\Rightarrow 4{\cos }^{2}A.{\cos }^{2}B+1-4\cos A.\cos B$ $=4{\sin }^{2}A.{\sin }^{2}B$

$\Rightarrow 4{\cos }^{2}A.{\cos }^{2}B-4\cos A.\cos B+1$ $=4\left[1-{\cos }^{2}A\right]\left[1-{\cos }^{2}B\right]$

$\Rightarrow 4{\cos }^{2}A.{\cos }^{2}B-4\cos A.\cos B+1$ $=4\left[1-{\cos }^{2}B-{\cos }^{2}A+{\cos }^{2}A.{\cos }^{2}B\right]$

$\Rightarrow 4{\cos }^{2}A.{\cos }^{2}B-4\cos A.\cos B+1=4-4{\cos }^{2}B-4{\cos }^{2}A+4{\cos }^{2}A.{\cos }^{2}B$

$\Rightarrow 4{\cos }^{2}B+4{\cos }^{2}A-4\cos A.\cos B=3$

$\therefore {\cos }^{2}B+{\cos }^{2}A-\cos A.\cos B=\frac{3}{4}$