Answer
(a) $y=2x$
(b) $y=-\frac{1}{2}x+\frac{5\pi}{8}$
Work Step by Step
To verify that the given point is on the curve, plug-in the coordinates to both sides of the equation: $LHS=\pi/4 sin(2\pi/2)=0$, $RHS=\pi/2 cos(2\pi/4)=0=LHS$
(a) Given the equation $x\cdot sin(2y)=y\cdot cos(2x)$, differentiate both sides with respect to $x$: $sin(2y)+2x\cdot cos(2y)y'=-2y\cdot sin(2x)+cos(2x)y'$ which gives $y'=\frac{sin(2y)+2y\cdot sin(2x)}{cos(2x)-2x\cdot cos(2y)}$
Thus the slope of the tangent line at the given point is $m=y'=\frac{sin(2\pi/2)+2(\pi/2)\cdot sin(2\pi/4)}{cos(2\pi/4)-2\pi/4\cdot cos(2\pi/2)}=\frac{0+\pi}{0+\pi/2}=2$ and the equation for the tangent line is $y-\frac{\pi}{2}=2(x-\frac{\pi}{4})$ or $y=2x$
(b) The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-\frac{1}{2}$, and the line equation is $y-\frac{\pi}{2}=-\frac{1}{2}(x-\frac{\pi}{4})$ or $y=-\frac{1}{2}x+\frac{5\pi}{8}$
