(a) (x²-4xy+4y²-4x²y²)/(x²y+2x²y²-2xy²)=
((x-2y)²-(2xy)²)/(xy(x+2xy-2y))=
(x-2y+2xy)(x-2y-2xy)/(xy(x-2y+2xy))=
(x-2y-2xy)/xy when x≠2 and y≠-1.
So as x→2, y→-1 (x-2y-2xy)/xy→(2+2+4)/-2=-4.
Therefore the limit is -4.
(b) Let h=π-x=π/2-y then sin(x)/cos(y)=sin(π-h)/cos(π/2-h)=
(sinπcos(h)-cosπsin(h))/(cos(π/2)cos(h)+sin(π/2)sin(h))=
sin(h)/sin(h)=1, because sinπ=cos(π/2)=0, cosπ=-1 and sin(π/2)=1.
Therefore the limit is 1 as h→0 (that is, x→π, y→π/2).