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分析（1）由已知得an+1=3an，（n≥2），从而得到数列{an}是以6为首项，公比为3的等比数列，由此能求出an．
（2）由$\frac{1}{{a}_{n}}$=$\frac{1}{2}•\frac{1}{{{3}^{n}}_{\;}}$，利用等比数列前n项和性质能求出数列{$\frac{1}{{a}_{n}}$}的前n项和．
解答解：（1）∵数列{an}的前n项和Tn满足an+1=2Tn+6，且a1=6，
∴由an+1=2Tn+6，①，得an=2Tn-1+6，（n≥2）②
②-①：有an+1-an=2Tn-2Tn-1，
即an+1=3an，（n≥2），
又a1=6，由②有a2=2T1+6=2a1+6=18，知a2=3a1，
∴数列{an}是以6为首项，公比为3的等比数列，
∴an=6•3n-1=2•3n．
（2）∵an=2•3n，∴$\frac{1}{{a}_{n}}$=$\frac{1}{2}•\frac{1}{{{3}^{n}}_{\;}}$，
∴数列{$\frac{1}{{a}_{n}}$}的前n项和：
Sn=$\frac{1}{2}$（$\frac{1}{3}+\frac{1}{{3}^{2}}+…+\frac{1}{{3}^{n}}$）=$\frac{1}{2}$×$\frac{\frac{1}{3}（1-\frac{1}{{3}^{n}}）}{1-\frac{1}{3}}$=$\frac{1}{4}-\frac{1}{4•{3}^{n}}$．
点评本题考查数列的通项公式和前n项和的求法，是中档题，解题时要认真审题，注意构造法和等比数列的性质的合理运用．
