1) Ta có \(\left. \begin{array}{l}
BC \bot AB\left( {ABCD\,\,\,\left( {hcn} \right)} \right)\\
BC \bot SA\left( {SA \bot \left( {ABCD} \right)} \right)
\end{array} \right\} \Rightarrow BC \bot \left( {SAB} \right) \Rightarrow BC \bot SB\)

Suy ra \(\Delta SBC\) vuông tại B

\(\left. \begin{array}{l}
CD \bot AD\left( {ABCD\,\,\,\left( {hcn} \right)} \right)\\
CD \bot SA\left( {SA \bot \left( {ABCD} \right)} \right)
\end{array} \right\} \Rightarrow CD \bot \left( {SAD} \right) \Rightarrow CD \bot SD\)

Suy ra \(\Delta SCD\) vuông tại D

2) Ta có

\(\begin{array}{l}
\left. \begin{array}{l}
CD\left( {SAD} \right)\\
AH\left( {SAD} \right)
\end{array} \right\} \Rightarrow CD \bot AH\\
\left. \begin{array}{l}
CD \bot AH\\
AH \bot SD
\end{array} \right\} \Rightarrow AH \bot \left( {SCD} \right)\\
 \Rightarrow AH \bot SC
\end{array}\)

3) 

\(\begin{array}{l}
\left. \begin{array}{l}
BC \bot \left( {SAB} \right)\\
AK \bot \left( {SAB} \right)
\end{array} \right\} \Rightarrow BC \bot AK\\
\left. \begin{array}{l}
BC \bot AK\\
AK \bot SB
\end{array} \right\} \Rightarrow AK \bot \left( {SBC} \right)\\
 \Rightarrow \left. \begin{array}{l}
AK \bot SC\\
AH \bot SC
\end{array} \right\} \Rightarrow SC \bot \left( {AHK} \right)\\
 \Rightarrow \left( {SAC} \right) \bot \left( {AHK} \right)
\end{array}\)

4) Dựng \(BC\bot AC\) tại I

\(BI\bot SA (SA\bot (ABCD)\)

\( \Rightarrow BI \bot \left( {SAC} \right)\) tại I

\( \Rightarrow \) SI là hình chiếu của SB trên (SAC)

\(\begin{array}{l}
 \Rightarrow \left( {\widehat {SB,\left( {SAC} \right)}} \right) = \left( {\widehat {SB,SI}} \right) = \widehat {BSI}\\
SB = \sqrt {S{A^2} + A{B^2}}  = a\sqrt 3 \\
\frac{1}{{B{I^2}}} = \frac{1}{{B{A^2}}} + \frac{1}{{B{C^2}}} \Rightarrow BI = \frac{{a\sqrt 3 }}{2}
\end{array}\)

\(\sin BSI = \frac{1}{2}\left( {\widehat {SB,\left( {SAC} \right)}} \right) = \widehat {BSI} = {30^0}\)