Đáp án:
$\begin{array}{l}
a)\overrightarrow {CB} = \overrightarrow {AB} - \overrightarrow {AC} = \left( { - 2; - 1;1} \right)\\
\Rightarrow \cos B = \frac{{\overrightarrow {AB} .\overrightarrow {CB} }}{{AB.CB}} = \frac{{2.\left( { - 2} \right) - 2.\left( { - 1} \right) - 4.1}}{{\sqrt {{2^2} + {2^2} + {4^2}} .\sqrt {{2^2} + 1 + 1} }} = - \frac{1}{2}\\
\Rightarrow \widehat B = {120^0}\\
b)\\
\overrightarrow {AM} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) = \left( {3; - \frac{3}{2}; - \frac{9}{2}} \right)\\
\Rightarrow AM = \sqrt {{3^2} + {{\left( { - \frac{3}{2}} \right)}^2} + {{\left( { - \frac{9}{2}} \right)}^2}} = \frac{{3\sqrt {14} }}{2}
\end{array}$