a) Ta có:

$\begin{cases}AB\bot BC\text{ (do ABC vuông tại B)}\\AB\bot SC\text{ (do SC vuông góc với đáy(ABC))}\end{cases}$

$\Rightarrow AB\bot(SBC)\Rightarrow AB\bot CK$

mà $CK\bot SB$

$\Rightarrow CK\bot(SAB)\Rightarrow CK\bot SA$ và $CK\bot HK\Rightarrow\Delta CHK\bot K$

và có $CH\bot SA$

$\Rightarrow SA\bot(CHK)$ (đpcm)

$\Rightarrow SA\bot HK$

b) $(SAC)\cap(SAB)=SA$

Mà $\Delta SAC$ có $CH\bot SA$

$\Delta SAB$ có: $HK\bot SA$

$\Rightarrow \widehat{((SAC),(SAB))}=(CH,HK)=\widehat{CHK}=\alpha$

$\Delta SAC\bot C$: $\dfrac{1}{CH^2}=\dfrac1{SC^2}+\dfrac{1}{AC^2}=\dfrac{1}{SC^2}+\dfrac1{3a^2}=\dfrac{3a^2+SC^2}{SC^2.3a^2}$

$\Rightarrow CH^2=\dfrac{SC^2.3a^2}{3a^2+SC^2}$

$\Delta SCB\bot C$: $\dfrac{1}{CK^2}=\dfrac{1}{SC^2}+\dfrac{1}{CB^2}=\dfrac{1}{SC^2}+\dfrac{1}{2a^2}=\dfrac{2a^2+SC^2}{SC^2.2a^2}$

$\Rightarrow CK^2=\dfrac{SC^2.2a^2}{2a^2+SC^2}$

$\Delta CHK\bot K$ có $\cos\alpha=\dfrac{HK}{CH}=\dfrac{\sqrt2}3$

$\Rightarrow \cos^2\alpha=\dfrac29$

$\Rightarrow \sin^2\alpha=79=\dfrac{CK^2}{CH^2}$

$\Rightarrow\dfrac{SC^2.2a^2}{2a^2+SC^2}:\dfrac{SC^2.3a^2}{3a^2+SC^2}=\dfrac79$

$\Rightarrow\dfrac{SC^2.2a^2}{2a^2+SC^2}.\dfrac{3a^2+SC^2}{SC^2.3a^2}=\dfrac79 $

$\Rightarrow \dfrac23.\dfrac{3a^2+SC^2}{2a^2+SC^2}=\dfrac79$

$\Rightarrow\dfrac{3a^2+SC^2}{2a^2+SC^2}=\dfrac76$

$\Rightarrow 1+\dfrac{a^2}{2a^2+SC^2}=\dfrac76$

$\Rightarrow\dfrac{a^2}{2a^2+SC^2}=\dfrac16$

$\Rightarrow 6a^2=2a^2+SC^2$

$\Rightarrow SC=2a$.