(1)由余弦定理：cosA=(b²+c²-a²)/2bc=bc/2bc=½;所以：A=π/3;
(2)2sin²B/2+2sin²C/2=1-cosB+1-cosC=1;cosB+cosC=1
B+C=π-A=2π/3;C=2π/3-B;
cosB+cos(2π/3-B)=1;cosB-½cosB+(√3/2)sinB =1;
½cosB+(√3/2)sinB=1;sin(B+π/6)=1;
所以B+π/6=π/2; B =π/3; C=π/3
所以三角形ABC是正三角形