$a)$ Ta có: $\left\{\begin{matrix} SA\perp BC\\ AB\perp BC \end{matrix}\right.\Rightarrow (SAB)\perp BC\Rightarrow SB\perp BC$
nên $SBC$ vuông tại $B.$
$b)$ $\left\{\begin{matrix} AC\perp BH\\ SA \perp BH\end{matrix}\right.\Rightarrow BH\perp (SAC)$
Mà $BH\subset (SBH)$ nên $(SAC)\perp (SBH)$
$c)$ Trong tam giác vuông $ABC,$ $BH$ là đường cao nên
$\frac{1}{BH^2}=\frac{1}{AB^2}+\frac{1}{BC^2}=\frac{1}{a^2}+\frac{1}{4a^2}=\frac{5}{4a^2}$
$\Rightarrow BH=\frac{2a}{\sqrt{5}}$
Trong tam giác vuông $BHC,$ $HC=\sqrt{BC^2-HB^2}=\frac{4a}{\sqrt{5}}$
$S_{\bigtriangleup BHC}=\frac{1}{2}BH.HC=\frac{4a^2}{5}$
$V_{SHBC}=\frac{1}{3}SA.S_{\bigtriangleup BHC}=\frac{1}{3}.3a.\frac{4a^2}{5}=\frac{4a^3}{5}$
Trong tam giác vuông $SAB:$ $SB=\sqrt{SA^2+AB^2}=a\sqrt{10}$
$S_{\bigtriangleup SBC}=\frac{1}{2}.SB.BC=a^2\sqrt{10}$
$V_{HSCB}=\frac{1}{3}.d(H,(SBC)).S_{\bigtriangleup SBC}=\frac{a^2.d(H,(SBC))\sqrt{10}}{3}$
Mà $V_{SHBC}=V_{HSCB}\Rightarrow \frac{4a^3}{5}=\frac{a^2.d(H,(SBC))\sqrt{10}}{3}$
$\Rightarrow d(H,(SBC))=\frac{12a}{5\sqrt{10}}$