解:∵根据正弦定理sinA/a=sinB/b=sinC/c=1/2R
又∵2asinA=(2b+c)sinB+(2c+b)sinC
∴2a^2=(2b+c)b+(2c+b)c
=2b^2+2c^2+2bc
∴b^2+c^2-a^2=-bc
即cosA=(b^2+c^2-a^2)/2bc=-1/2
A=120°......
由正弦定理:2a^2=2b^2+bc+2c^2+bc
a^2=b^2+c^2+bc
bc=-2bccosA
cosA=-1/2
A=120度
sinB+sinC=2sin(B+C)/2*cos(B-C)/2=2sin30度*cos(2B-60度)/2[和差化积]
B=30度,cos(2B-60度)=1
sinB+sinC最大2sin30度=1