1.∵△ABC是锐角三角形
∴有（1）C=π-(A+B)
（2）sinA+cosA>0
（3）若sinB=cosB,则B=π/4
∵sin(A-B)=cosC
==>sin(A-B)=cos[π-(A+B)] (应用(1))
==>sin(A-B)=-cos(A+B)
==>sinAcosB-cosAsinB=-cosAcosB+sinAsinB (应用和差角公式)
==>sinAsinB+cosAsinB=sinAcosB+cosAcosB
==>sinB(sinA+cosA)=cosB(sinA+cosA)
==>sinB=cosB (应用(2))
∴B=π/4 (应用(3)).
2.∵a=3√2,b=√10
∴由余弦定理,得
c^2+(3√2)^2-6√2c*cos(π/4)=(3√2)^2
==>c^2-6c+8=0
==>(c-2)(c-4)=0
故c=2,或c=4.