Gọi H là trung điểm của AB, vì tam giac SAB là tam giác đều nên SH ^ AB.

Ta có: \(\left\{ \begin{array}{l}\left( {SAB} \right) \cap \left( {ABC} \right) = AB\\\left( {SAB} \right) \bot \left( {ABC} \right)\\\left( {SAB} \right) \supset SH \bot AB\end{array} \right. \Rightarrow SH \bot \left( {ABC} \right)\)

Tam giác SAB đều cạnh 2a nên

\(SH = \frac{{2a\sqrt 3 }}{2} = a\sqrt 3 \)

\({S_{ABC}} = \frac{1}{2}\,.\,2a\,.\,2a = 2{a^2}\)

\( \Rightarrow {V_{S.ABC}} = \frac{1}{2}SH\,.\,{S_{ABC}} = \frac{1}{3}\,.\,a\sqrt 3 \,.\,2{a^2} = \frac{{2{a^3}\sqrt 3 }}{3}\)

Lại có:

\(\frac{{{V_{S.AMN}}}}{{{V_{S.ABC}}}} = \frac\,.\,\frac = \frac{1}{2}\,.\,\frac{1}{3} = \frac{1}{6}\)

\( \Rightarrow {V_{S.AMN}} = \frac{1}{6}{V_{S.ABC}} = \frac{1}{6}\,.\,\frac{{2{a^3}\sqrt 3 }}{3} = \frac{{{a^3}\sqrt 3 }}{9}\).