解析：∵3b^2+3c^2-3a^2=4√2bc,
∴(b^2+c^2-a^2)/2bc=2√2/3
即cosA=2√2/3
sinA=√[1-(cosA)^2]=√(1-8/9)=1/3
2sin（A+π/4）=√2（sinA+cosA)
=√2(1/3+2√2/3)=√2/3+4/3
sin（B+C+π/4）/(1-cos2A)
=sin(A-π/4）/(1-cos2A)
=√2/2(sinA-cosA）/2(sinA)^2
=√2/2(1/3-2√2/3)/(2/9)
=3√2/4-3