\[\begin{array}{l} 1)\,\,\,Xet\,\,\Delta ABK\,\,\,vuong\,\,tai\,\,K\\ \Rightarrow AK = AB\cos \angle BAC = \sqrt 2 .\cos {60^0} = \frac{{\sqrt 2 }}{2}.\\ BK = AB\sin \angle BAC = \sqrt 2 \sin {60^0} = \frac{{\sqrt 6 }}{2}.\\ \angle ABC = {180^0} - \angle ACB - \angle BAC = {180^0} - {45^0} - {60^0} = {75^0}.\\ Xet\,\,\Delta AHC\,\,vuong\,\,tai\,\,H\,\,co:\\ BC = \frac{{BK}}{{\cos \angle C}} = \frac{{\sqrt 6 }}{{2.\frac{{\sqrt 2 }}{2}}} = \sqrt 3 .\\ KC = BC.\cos \angle C = \sqrt 3 .\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 }}{2}.\\ \Rightarrow AC = AK + KC = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 6 }}{2} = \frac{{\sqrt 6 + \sqrt 2 }}{2}.\\ Xet\,\,\Delta AHC\,\,vuong\,\,tai\,\,H\,\,co:\\ AH = AC.sin{45^0} = \frac{{\sqrt 6 + \sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 3 + 1}}{2}.\\ 2)\,\,Xet\,\,\Delta ABH\,\,\,vuong\,\,tai\,\,H\,\,\,ta\,\,co:\\ \sin {75^0} = \sin \angle B = \cos {15^0} = \frac{{AH}}{{AB}} = \frac{{\sqrt 3 + 1}}{{2.\sqrt 2 }} = \frac{{\sqrt 6 + \sqrt 2 }}{4}.\\ \cos {75^0} = \cos \angle B = \sin {15^0} = \sqrt {1 - {{\sin }^2}{{75}^0}} = \sqrt {1 - {{\left( {\frac{{\sqrt 6 + \sqrt 2 }}{4}} \right)}^2}} = \frac{{\sqrt 6 - \sqrt 2 }}{4}. \end{array}\]

1) Áp dụng hệ thứ lượng vào tam giác vuông $ABK$ ta có:

$\cos\widehat{BAK}=\dfrac{AK}{AB}$

$\Rightarrow AK=AB.\cos\widehat{BAK}=\sqrt2.\cos60^o=\dfrac{1}{\sqrt2}$.

$\sin\widehat{BAK}=\dfrac{BK}{AB}$

$\Rightarrow BK=AB.\sin\widehat{BAK}=\sqrt2.\sin60^o=\dfrac{\sqrt3}{\sqrt2}$.

Áp dụng hệ thứ lượng vào tam giác vuông $CBK$ ta có:

$\tan\widehat{BCK}=\dfrac{BK}{KC}$

$\Rightarrow KC=\dfrac{BK}{\tan\widehat{BAK}}=\dfrac{\dfrac{\sqrt3}{\sqrt2}}{\tan45^o}=\dfrac{\sqrt3}{\sqrt2}$

$\cos\widehat{BCK}=\dfrac{KC}{BC}$

$\Rightarrow BC=\dfrac{KC}{\cos\widehat{BCK}}=\dfrac{\dfrac{\sqrt3}{\sqrt2}}{\cos45^o}=\dfrac{\sqrt3}{2}$.

$AC=AK+KC=\dfrac{1}{\sqrt2}+\dfrac{\sqrt3}{\sqrt2}=\dfrac{1+\sqrt3}{\sqrt2}$

Áp đụng hệ thức lượng vào tam giác vuông $AHC$:

$\cos\widehat{HCA}=\dfrac{AH}{AC}$

$\Rightarrow AH=AC.\cos\widehat{HCA}=\dfrac{1+\sqrt3}{\sqrt2}\cos 45^0=\dfrac{1+\sqrt3}{2}$.