1.设三角形ABC的内角A.B.C得对边长分别为a,b,c,cos(A-C)+cosB=3/2,

1.
b的平方=ac
正弦：sin²B=sinAsinC
cos(A-C)+cosB=3/2 cosB=cos(π-(A+C))=-cos(A+C)
cosAcosC+sinAsinC-cosAcosC+sinAsinC=3/2
2sinAsinC=3/2
sin²B=3/4
sinB属于(0,1〕
所以sinB=√3/2
2.
tanC=sinA+sinB/cosA+cosB,
c/cosC=(a+b)/(cosA+cosB)
(b&sup2+c&sup2-a&sup2)/2b +(a&sup2+c&sup2-b&sup2)/2a
=(a+b)(a&sup2+b&sup2-c&sup2)/2ab
ab&sup2+ac&sup2-a&sup3+ba&sup2+bc&sup2-b&sup3
=a&sup3+ab&sup2-ac&sup2+ba&sup2+b&sup3-bc&sup2
2ac&sup2+2bc&sup2=2a&sup3+2b&sup3
(a+b)c&sup2=(a+b)(a&sup2-ab+b&sup2)
c&sup2=a&sup2+b&sup2-ab
ab=a&sup2+b&sup2-c&sup2
因为：
2abcosC=a&sup2+b&sup2-c&sup2
所以：
2cosC=1
cosC=1/2
C=π/3 0＜B,A＜2π/3
sin(B-A）=cosC=1/2
B-A=π/6,5π/6（舍）
因为B+A=2π/3
所以A=π/4
B=5π/12
2）三角形ABC的面积=3+根号3
ac/2 sin(π/4+π/6）=3+√3
a/sinπ/4=c/sinπ/3
自己解吧