$\begin{array}{l} 1)MB = 2MC\\ \Rightarrow 2\overrightarrow {MC} = \overrightarrow {BM} \\ \Rightarrow 3\overrightarrow {MC} = \overrightarrow {BM} + \overrightarrow {MC} = \overrightarrow {BC} \\ \overrightarrow {AM} = \overrightarrow {AC} + \overrightarrow {CM} = \overrightarrow {AC} - \dfrac{1}{3}\overrightarrow {BC} = \overrightarrow {AC} - \dfrac{1}{3}\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)\\ = \dfrac{2}{3}\overrightarrow {AC} - \dfrac{1}{3}\overrightarrow {AB} \\ 2)\overrightarrow {AM} = \overrightarrow {AB} + \overrightarrow {BM} = \overrightarrow {AB} + 3\overrightarrow {CM} = \overrightarrow {AB} + 3\left( {\overrightarrow {AM} - \overrightarrow {AC} } \right)\\ \Leftrightarrow - 2\overrightarrow {AM} = \overrightarrow {AB} + 3\overrightarrow {AC} \\ \Leftrightarrow \overrightarrow {AM} = - \dfrac{1}{2}\overrightarrow {AB} + \dfrac{3}{2}\overrightarrow {AC} = - \dfrac{1}{2}\overrightarrow u + \dfrac{3}{2}\overrightarrow v \\ 3)MA = MB\left( {gt} \right) \Rightarrow \overrightarrow {MA} = - \overrightarrow {MB} = - \dfrac{1}{2}\overrightarrow {AB} \\ \Rightarrow \overrightarrow {AM} = \dfrac{1}{2}\overrightarrow {AB} \\ NC = 2NA \Rightarrow \overrightarrow {NC} = - 2\overrightarrow {NA} \Rightarrow \overrightarrow {NA} = - \dfrac{1}{3}\overrightarrow {AC} \\ \Rightarrow \overrightarrow {AN} = \dfrac{1}{3}\overrightarrow {AC} \\ a)\overrightarrow {AK} = \dfrac{1}{2}\left( {\overrightarrow {AM} + \overrightarrow {AN} } \right) = \dfrac{1}{2}\left( {\dfrac{1}{2}\overrightarrow {AB} + \dfrac{1}{3}\overrightarrow {AC} } \right)\\ = \dfrac{1}{4}\overrightarrow {AB} + \dfrac{1}{6}\overrightarrow {AC} \\ b)\overrightarrow {KD} = \overrightarrow {AD} - \overrightarrow {AK} = \dfrac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) - \dfrac{1}{4}\overrightarrow {AB} - \dfrac{1}{6}\overrightarrow {AC} \\ = \dfrac{1}{4}\overrightarrow {AB} + \dfrac{1}{3}\overrightarrow {AC} \end{array}$
