Đáp án:

$\begin{array}{l}
1)Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
 \Rightarrow {15^0} + {20^2} = B{C^2}\\
 \Rightarrow B{C^2} = 625\\
 \Rightarrow BC = 25\left( {cm} \right)\\
 \Rightarrow AH = \dfrac{{AB.AC}}{{BC}} = \dfrac{{15.20}}{{25}} = 12\left( {cm} \right)\\
2)\\
AM = \dfrac{1}{2}BC \Rightarrow BC = 2.AM = 6\\
 \Rightarrow AC = \sqrt {B{C^2} - A{B^2}}  = \sqrt {{6^2} - 2}  = \sqrt {34} \\
3){b^2} + {c^2} - {a^2} = 2.b.c.cos\widehat A\\
 \Rightarrow {8^2} + {5^2} - {a^2} = 2.8.5.\cos {60^0}\\
 \Rightarrow {a^2} = 49\\
 \Rightarrow a = 7\\
 \Rightarrow m_a^2 = \dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4} = \dfrac{{129}}{4}\\
 \Rightarrow {m_a} = \dfrac{{\sqrt {129} }}{2}\\
Do:2R = \dfrac{a}{{\sin \widehat A}}\\
 \Rightarrow 2R = \dfrac{7}{{\sin {{60}^0}}} \Rightarrow R = \dfrac{{7\sqrt 3 }}{3}\\
S = \dfrac{{abc}}{{4r}} = \dfrac{1}{2}.b.c.\sin \widehat A = \dfrac{1}{2}.8.5.\sin {60^0}\\
 \Rightarrow S = \dfrac{{8.5.7}}{{4r}} = 10\sqrt 3 \\
 \Rightarrow r = \dfrac{{7\sqrt 3 }}{3}\\
 \Rightarrow {h_a} = \dfrac{{2S}}{a} = \dfrac{{20\sqrt 3 }}{7}\\
4)a)p = \dfrac{{a + b + c}}{2} = 13\\
 \Rightarrow S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \\
 = \sqrt {13.\left( {13 - 7} \right)\left( {13 - 9} \right)\left( {13 - 10} \right)} \\
 = 6\sqrt {26} \\
r = \dfrac{{abc}}{{4S}} = 5,15\\
\cos \widehat A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2.b.c}} = \dfrac{{{9^2} + {{10}^2} - {7^2}}}{{2.9.10}} = \dfrac{{11}}{{15}}\\
B6)\\
B{C^2} + A{C^2} - A{B^2} = 2.BC.AC.cos\widehat C\\
 \Rightarrow 2 + A{C^2} - 3 = 2.\sqrt 2 .AC.cos{60^0}\\
 \Rightarrow A{C^2} - \sqrt 2 .AC - 1 = 0\\
 \Rightarrow AC = \dfrac{{\sqrt 6  + \sqrt 2 }}{2}
\end{array}$