Đáp án:

$\begin{array}{l}
a)\overrightarrow {AB}  =  - \frac{2}{3}\overrightarrow {CM}  - \frac{4}{3}\overrightarrow {BN} \\
 \Rightarrow \overrightarrow {AB}  =  - \frac{2}{3}.\frac{1}{2}\left( {\overrightarrow {CA}  + \overrightarrow {CB} } \right) - \frac{4}{3}.\frac{1}{2}\left( {\overrightarrow {BA}  + \overrightarrow {BC} } \right)\\
 \Rightarrow \overrightarrow {AB}  =  - \frac{1}{3}.\overrightarrow {CA}  - \frac{1}{3}\overrightarrow {CB}  - \frac{2}{3}\overrightarrow {BA}  - \frac{2}{3}\overrightarrow {BC} \\
 \Rightarrow \overrightarrow {AB}  = \frac{1}{3}\overrightarrow {AC}  + \frac{1}{3}\overrightarrow {BC}  + \frac{2}{3}\overrightarrow {AB}  - \frac{2}{3}\overrightarrow {BC} \\
 \Rightarrow \frac{1}{3}\overrightarrow {AB}  - \frac{1}{3}\overrightarrow {AC}  = \frac{{ - 1}}{3}\overrightarrow {BC} \\
 \Rightarrow \frac{1}{3}\overrightarrow {CB}  = \frac{1}{3}\overrightarrow {CB} \left( {ld} \right)\\
b)\\
\overrightarrow {AC}  =  - \frac{4}{3}\overrightarrow {CM}  - \frac{2}{3}\overrightarrow {BN} \\
 \Rightarrow \overrightarrow {AC}  =  - \frac{4}{3}.\frac{1}{2}\left( {\overrightarrow {CA}  + \overrightarrow {CB} } \right) - \frac{2}{3}.\frac{1}{2}\left( {\overrightarrow {BA}  + \overrightarrow {BC} } \right)\\
 \Rightarrow \overrightarrow {AC}  =  - \frac{2}{3}\overrightarrow {CA}  - \frac{2}{3}\overrightarrow {CB}  - \frac{1}{3}.\overrightarrow {BA}  - \frac{1}{3}\overrightarrow {BC} \\
 \Rightarrow \overrightarrow {AC}  + \frac{2}{3}\overrightarrow {CA}  = \frac{2}{3}\overrightarrow {BC}  - \frac{1}{3}\overrightarrow {BC}  + \frac{1}{3}\overrightarrow {AB} \\
 \Rightarrow \frac{1}{3}\overrightarrow {AC}  = \frac{1}{3}\overrightarrow {BC}  + \frac{1}{3}\overrightarrow {AB}  = \frac{1}{3}\overrightarrow {AC} \left( {ld} \right)\\
C)\overrightarrow {MN}  = \frac{1}{3}\overrightarrow {BN}  - \frac{1}{3}\overrightarrow {CM} \\
 \Rightarrow \frac{1}{2}\overrightarrow {BC}  = \frac{1}{3}.\frac{1}{2}\left( {\overrightarrow {BA}  + \overrightarrow {BC} } \right) - \frac{1}{3}.\frac{1}{2}\left( {\overrightarrow {CA}  + \overrightarrow {CB} } \right)\\
 \Rightarrow \frac{1}{2}\overrightarrow {BC}  = \frac{1}{6}\overrightarrow {BA}  + \frac{1}{6}\overrightarrow {BC}  - \frac{1}{6}\overrightarrow {CA}  - \frac{1}{6}\overrightarrow {CB} \\
 \Rightarrow \frac{1}{2}\overrightarrow {BC}  - \frac{1}{6}\overrightarrow {BC}  - \frac{1}{6}\overrightarrow {BC}  = \frac{1}{6}\left( {\overrightarrow {BA}  + \overrightarrow {AC} } \right)\\
 \Rightarrow \frac{1}{6}\overrightarrow {BC}  = \frac{1}{6}\overrightarrow {BC} \left( {ld} \right)
\end{array}$