1. Ta có \(\angle DAB = {120^0} \Rightarrow \angle DAH = {30^0}.\)

Vì \(ABCD\) là hình thang cân \( \Rightarrow DH = \left( {CD - AB} \right):2 = \left( {2a - a} \right):2 = \frac{a}{2}.\)

Xét tam giác ADH vuông tại H ta có:

\(\begin{array}{l}AD = \frac{{AH}}{{\sin {{30}^0}}} = \frac{a}{{2.\frac{1}{2}}} = a.\\AH = \sqrt {A{D^2} - D{H^2}}  = \sqrt {{a^2} - \frac{{{a^2}}}{4}}  = \frac{{a\sqrt 3 }}{2}.\\\overrightarrow {AH} .\left( {\overrightarrow {CD}  - 4\overrightarrow {AD} } \right) = \overrightarrow {AH} .\overrightarrow {CD}  - 4\overrightarrow {AH} .\overrightarrow {AD} \\ = AH.CD.\cos \left( {\overrightarrow {AH} ,\,\,\overrightarrow {CD} } \right) - 4AH.AD.\cos \left( {\overrightarrow {AH} ,\,\,\overrightarrow {AD} } \right)\\ = AH.CD.\cos {90^0} - 4AH.AD.\cos {30^0}\\ =  - 4.\frac{{a\sqrt 3 }}{2}.a.\frac{{\sqrt 3 }}{2} =  - 3{a^2}.\\\overrightarrow {AC} .\overrightarrow {BH}  = \left( {\overrightarrow {AB}  + \overrightarrow {BC} } \right).\left( {\overrightarrow {BC}  + \overrightarrow {CH} } \right)\\ = \overrightarrow {AB} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CH}  + B{C^2} + \overrightarrow {BC} .\overrightarrow {CH} \\ = AB.BC.\cos {60^0} + AB.CH.\cos {90^0} + B{C^2} + BC.CH.\cos {120^0}\\ = a.a.\frac{1}{2} + {a^2} + a.\frac{{3a}}{2}.\left( { - \frac{1}{2}} \right)\\ = \frac{{{a^2}}}{2} + {a^2} - \frac{{3{a^2}}}{4} = \frac{{3{a^2}}}{4}.\end{array}\)

2. \(A\left( {2; - 3} \right),\,\,\,B\left( {1; - 2} \right) \Rightarrow \overrightarrow {AB} = \left( { - 1;\,\,1} \right).\)

a) \(\overrightarrow u = 3\overrightarrow i - 3\overrightarrow j  \Rightarrow \overrightarrow u  = \left( {3; - 3} \right) =  - 3\left( { - 1;\,\,1} \right).\)  

Ta thấy:\(\overrightarrow u  =  - 3\overrightarrow {AB}  \Rightarrow \overrightarrow u ,\,\,\,\overrightarrow {AB} \) cùng phương.

Ta có: \(\left\{ \begin{array}{l}\left| {\overrightarrow {AB} } \right| = \sqrt 2 \\\left| {\overrightarrow u } \right| = 3\sqrt 2 \end{array} \right.\) \( \Rightarrow k = \left| {\overrightarrow {AB} } \right|:\left| {\overrightarrow u } \right| = \sqrt 2 :3\sqrt 2  = \frac{1}{3}.\)