解：（1）a=bsin(π／4+C)-csin(π／4+B)=bsin(A+C)-csin(A+B)=bsinB-csinC ①由正弦定理a/sinA=b/sinB=c/sinC=2R (R为三角形ABC外接圆半径)得a=2RsinA,b=2RsinB,c=2RsinC代入①式可得2RsinA=2R(sinB)^2-2R(sinC)^2于是sinA=(sinB)^2-(sinC)^2=(1-cos2B)/2-(1-cos2C)/2=-(cos2B-cos2C)/2=sin(B+C)sin(B-C)=sinAsin(B-C)因sinA≠0，故有sin(B-C)=1，则B-C=π/2（2）由正弦定理得2R=a/sinA=√2/(sinπ/4)=2故b=2RsinB,c=2RsinC故S△ABC=1/2*bcsinA=1/2*2RsinB*2RsinC*sin(π/4)=1/2*2sinB*2sinC*√2/2=√2/2*2sinBsinC=√2/2*[cos(B-C)-cos(B+C)]=√2/2*[0-cos(3π/4)]=1/2其中B+C=π-A=3π/4。
