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(B) Given $\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$.Using the identity $\cos^2 B - \sin^2 A = \cos(A+B) \cos(A-B)$,we have $\cos(A+B) \cos(

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$\frac{1}{2}$

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(B) Given $\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$.Using the identity $\cos^2 B - \sin^2 A = \cos(A+B) \cos(A-B)$,we have $\cos(A+B) \cos(A-B) = \frac{\sqrt{3}+1}{4\sqrt{2}} \dots (i)$.Given $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$.Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $\cos(A+B) + \cos(A-B) = \frac{1+\sqrt{3}}{2\sqrt{2}} + \frac{1}{2} \dots (ii)$.Let $x = \cos(A+B)$ and $y = \cos(A-B)$. From $(i)$ and $(ii)$,$xy = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $x+y = \frac{\sqrt{3}+1}{2\sqrt{2}} + \frac{1}{2}$.Solving these,we get $x = \frac{1}{2} = \cos 60^{\circ}$ and $y = \frac{\sqrt{3}+1}{2\sqrt{2}} = \cos 15^{\circ}$.Thus,$A+B = 60^{\circ}$ and $A-B = 15^{\circ}$.Solving for $A$ and $B$,we get $2A = 75^{\circ} \implies A = 37.5^{\circ}$ and $2B = 45^{\circ} \implies B = 22.5^{\circ}$.Now,$\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} = \cos^2 \frac{4(22.5^{\circ})}{3} - \sin^2 \frac{4(37.5^{\circ})}{5} = \cos^2 30^{\circ} - \sin^2 30^{\circ} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

If $0 < B < A < \frac{\pi}{4}$,$\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$,then $\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} =$

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