a) Ta có: \(\overrightarrow {AB} .\overrightarrow {AC} = AB.AC.\cos \widehat {BAC} = a.a.\cos 60^\circ = \frac{{{a^2}}}{2};\)

\(\overrightarrow {BC} .\overrightarrow {AC} = \overrightarrow {CB} .\overrightarrow {CA} = CB.CA.\cos \widehat {BCA} = a.a.\cos 60^\circ = \frac{{{a^2}}}{2}.\)

b) Ta có: \(3\overrightarrow {BM} = 2\overrightarrow {BC} \)

⇔ \(3\left( {\overrightarrow {AM} - \overrightarrow {AB} } \right) = 2\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)\)

⇔ \(\overrightarrow {AM} = \frac{2}{3}\overrightarrow {AC} + \frac{1}{3}\overrightarrow {AB} \)

Ta có: \(5\overrightarrow {AN} = 4\overrightarrow {AC} \)

⇔ \(5\left( {\overrightarrow {BN} - \overrightarrow {BA} } \right) = 4\overrightarrow {AC} \)

⇔ \(\overrightarrow {BN} = - \overrightarrow {AB} + \frac{4}{5}\overrightarrow {AC} \)

Ta có: \(\overrightarrow {AM} .\overrightarrow {BN} = \left( {\frac{2}{3}\overrightarrow {AC} + \frac{1}{3}\overrightarrow {AB} } \right)\left( { - \overrightarrow {AB} + \frac{4}{5}\overrightarrow {AC} } \right)\)

\( = - \frac{2}{3}\overrightarrow {AC} .\overrightarrow {AB} + \frac{8}{{15}}{\overrightarrow {AC} ^2} - \frac{1}{3}{\overrightarrow {AB} ^2} + \frac{4}{{15}}\overrightarrow {AC} .\overrightarrow {AB} \)

\( = - \frac{2}{5}\overrightarrow {AC} .\overrightarrow {AB} + \frac{8}{{15}}{\overrightarrow {AC} ^2} - \frac{1}{3}{\overrightarrow {AB} ^2}\)

\( = - \frac{2}{3}.\frac{{{a^2}}}{2} + \frac{8}{{15}}{a^2} - \frac{1}{3}{a^2} = 0\)

⇒ AM ⊥ BN ⇒ AM vuông góc với BN.