1.a^2=b^2+c^2+√3bc,
cosA=-√3/2,
A=150°,sinA=1/2,
a=√3,外接圆半径R=a/(2sinA)=√3,
S+3cosBcosC
=（1/2）bcsinA+3cosBcosC
=sinBsinC+3cosBcosC
=cos(B-C)+2cosBcosC
=cos(B+C)+2cos(B-C)
=-cosA+2cos(B-C)
=√3/2+2cos(B-C),
它的最大值=√3/2+2.
2.a1=25,a11^2=a1a13.
∴(25+10d)^=25(25+12d),
500d+100d^=300d,d≠0,
∴d=-2,
∴an=25+(n-1)(-2)=27-2n.
a=27-2(3n-2)=31-6n,
S=a1+a4+a7+...+a=n(25+31-6n)/2=n(28-3n).
3.4Sn=a^2-4n-1,n属于N*,①
令n=1得4a1=a2^-4-1,
∴a2^=4a1+5,a2>0,
∴a2=√(4a1+5).
令n=2,得4[a1+√(4a1+5)]=a3^-9,
a3^=4a1+4√(4a1+5)+9=[2+√(4a1+5)]^,a3>0,
∴a3=2+a2,
令n=3,得4（a1+a2+a3)=a4^-13,
a4^=4a1+8a2+21=(a2+4)^,a4=a2+4=a3+2,
假设a=a2+2(k-2),则
Sk=a1+(k-1)[a2+a]/2=a1+(k-1)(a2+k-2),
由①,4[a1+(k-1)(a2+k-2)]=a^-4k-1,
a^=4a1+4(k-1)a2+4(k^-3k+2)+4k+1,
=(4a1+5)+4(k-1)a2+4k^-8k+4
=a2^+4(k-1)a2+(2k-2)^
=(a2+2k-2)^,
a>0,
∴a=a2+2k-2=a2+2(k-1),
综上,an={a1,
{[√(4a1+5)+2(n-2)],n>=2.