(1)sin(π/4+a)=cos[π/2-(π/4+a)]=cos(π/4-a)=3/5；
cos(π/4+b)=sin[π/2+(π/4+b)]=sin(3π/4+b)=5/13,.
(2)因为π/4所以-π/2<π/4-a<0,π/4<π/4+b<π/2,
因此sin(π/4-a)=-√[1-(3/5)^2]=-4/5,sin(π/4+b)=√[1-(5/13)^2]=12/13,
所以cos(a+b)=cos[(π/4+b)-(π/4-a)]
=cos(π/4+b)cos(π/4-a)+sin(π/4+b)sin(π/4-a)
=(5/13)*(3/5)+(12/13)*(-4/5)=-33/65