a) Ta có: \(\overrightarrow {EF} = \overrightarrow {EA} + \overrightarrow {AD} + \overrightarrow {DF} \)

                     = −\(\overrightarrow {AE} \) + \(\overrightarrow {AD} \) − \(\overrightarrow {FD} \)

                     = \(\overrightarrow {AD} \)− \(\frac{1}{3}\overrightarrow {AB} \) − \(\frac{2}{3}\overrightarrow {CD} \).

Vậy \(\overrightarrow {EF} = \overrightarrow {AD} - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {CD} \).

b) Ta có: \(\overrightarrow {EF} = \overrightarrow {EB} + \overrightarrow {BC} + \overrightarrow {CF} \)

                     = \(\frac{2}{3}\overrightarrow {AB} + \overrightarrow {CB} + \frac{1}{3}\overrightarrow {CD} \).

Vậy \(\overrightarrow {EF} = \overrightarrow {BC} + \frac{2}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {CD} \).

c) Từ câu a và b, ta có:

\(3\overrightarrow {EF} = \left( {\overrightarrow {AD} - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {CD} } \right) + 2\left( {\frac{2}{3}\overrightarrow {AB} + \overrightarrow {CB} + \frac{1}{3}\overrightarrow {CD} } \right)\)

        = \(\overrightarrow {AD} - \frac{1}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {CD} \) + \(\frac{4}{3}\overrightarrow {AB} + 2\overrightarrow {CB} + \frac{2}{3}\overrightarrow {CD} \)

        = \(\overrightarrow {AD} \) + \(\left( { - \frac{1}{3}\overrightarrow {AB} + \frac{4}{3}\overrightarrow {AB} } \right)\) + \(\left( { - \frac{2}{3}\overrightarrow {CD} + \frac{2}{3}\overrightarrow {CD} } \right)\) + \(2\overrightarrow {CB} \)

        = \(\overrightarrow {AD} \) + \(2\overrightarrow {CB} \) + \(\overrightarrow {AB} \)

⇒ \(\overrightarrow {EF} = \frac{1}{3}\overrightarrow {AD} + \frac{2}{3}\overrightarrow {BC} + \frac{1}{3}\overrightarrow {AB} \).

Vậy ta có đpcm.