1 （1）由正弦定理:-b/(2a+c)=-sinB/(2sinA+sinC)
故cosB/cosC=-sinB/(2sinA+sinC)
-sinBcosC=2sinAcosB+cosBsinC
既-(sinBcosC+cosBsinC)=2sinAcosB
-sin(B+C)=2sinAcosB
又B+C=180-A 故sin(B+C)=sinA
-sinA=2sinAcosB cosB=-1/2 B=120度
（2）(a+c)^2=16
既a^2+c^2+2ac=16
又由余弦定理a^2+c^2-b^2=2accosB
既16-2ac-13=-ac ac=3
S=1/2acsinB=3√3/4
2 (1)由題意b^2=ac
由正弦定理,可化為(sinB)^2=sinAsinC
又cosB=3/4 故sinB=√7/4
sinAsinC=7/16
1/tanA+1/tanC=cosA/sinA+cosC/sinC
=(cosAsinC+sinAcosC）/sinAsinC
=sin(A+C)/sinAsinC
又A+C=180-B 故sin(A+C)=sinB
既1/tanA+1/tanC=(√7/4)/(7/16)=4√7/7
(2)向量BA*向量BC(數(shù)量積)
=|BA|*|BC|*cosB
=c*acosB=3/2
故ac*(3/4)=3/2 ac=2
再由余弦定理
a^2+c^2-b^2=2accosB
且b^2=ac
故a^2+c^2-ac=2accosB
a^2+c^2-2=3
a^2+c^2=5
(a+c)^2=a^2+c^2+2ac=5+4=9
故a+c=3