由ABC成等差数列,则有2B=A+C,又A+B+C=π,所以,B=π/3
①由正弦定理得：a/sinA=b/sinB=c/sinC,所以,sinA=asinB/b=3×(√3/2)/√13=3√39/26
由于,a=3＜b=√13,所以A＜B=π/3,则,cosA=√（1-sin²A）=5√13/26
sinC=sin（π-A-B）=sin（A+B）=sinAcosB+cosAsinB
=（3√39/26）×（1/2）+5√13/26×（√3/2）
=7√39/52
所以,c=bsinC/sinB=√13×（7√39/52）/（√3/2）=7/2
②t=sinAsinC =sin（π-B-C）sinC=sin(2π/3-C)sinC
=(sin2π/3×cosC-cos2π/3×sinC)sinC
=（√3/2）sinCcosC-(1/2）sin²C
=（√3/4）sin2C-(1/4）(1-cos2C)
=（√3/4）sin2C+(1/4）cos2C-1/4
=（1/8）sin（2C+π/6）-1/4
由于,0＜C＜2π/3,所以,π/6＜2C+π/6＜3π/2
所以当sin（2C+π/6）=sin（π/2）=1时,t取得最大值
即t=1/8-1/4=-1/8