a) Ta có: \(\left\{ {\begin{array}{*{20}{l}}{BC \bot AB{\mkern 1mu} {\mkern 1mu} \left( {gt} \right)}\\{BC \bot SA{\mkern 1mu} {\mkern 1mu} \left( {SA \bot \left( {ABCD} \right)} \right)}\end{array}} \right. \Rightarrow BC \bot \left( {SAB} \right) \Rightarrow \left( {SBC} \right) \bot \left( {SAB} \right)\).

\(\left\{ {\begin{array}{*{20}{l}}{BD \bot SA{\mkern 1mu} {\mkern 1mu} \left( {SA \bot \left( {ABCD} \right)} \right)}\\{BD \bot AC{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {gt} \right)}\end{array}} \right. \Rightarrow BD \bot \left( {SAC} \right) \Rightarrow \left( {SBD} \right) \bot \left( {SAC} \right)\).

b) Ta có \(BC \bot \left( {SAB} \right){\mkern 1mu} {\mkern 1mu} \left( {cmt} \right) \Rightarrow \widehat {\left( {SC;\left( {SAB} \right)} \right)} = \widehat {\left( {SC;SB} \right)} = \widehat {CSB}\).

Trong tam giác SBC vuông tại \(B\) ta có :

\(\begin{array}{*{20}{l}}{SB = \sqrt {S{A^2} + A{B^2}} {\rm{\;}} = \sqrt {12{a^2} + 4{a^2}} {\rm{\;}} = 4a}\\{BC = 2a \Rightarrow \tan \widehat {CSB} = \frac{{BC}}{{SB}} = \frac{1}{2} \Rightarrow \widehat {CSB} \approx 26^\circ 34'}\end{array}\)

Vậy \(\widehat {\left( {SC;\left( {SAB} \right)} \right)} \approx 26^\circ 34'\).

c) Gọi \(O = AC \cap BD\). Ta có: \(BD \bot \left( {SAC} \right){\mkern 1mu} {\mkern 1mu} \left( {cmt} \right) \Rightarrow BD \bot SO\).

\(\left\{ {\begin{array}{*{20}{l}}{\left( {SBD} \right) \cap \left( {ABCD} \right) = BD}\\{\left( {SBD} \right) \supset SO \bot BD}\\{\left( {ABCD} \right) \supset AO \bot BD}\end{array}} \right. \Rightarrow \widehat {\left( {\left( {SBD} \right);\left( {ABCD} \right)} \right)} = \widehat {\left( {SO;AO} \right)} = \widehat {SOA}\).

ABCD là hình vuông cạnh \(2a \Rightarrow AC = BD = 2a\sqrt 2 {\rm{\;}} \Rightarrow AO = \frac{1}{2}AC = a\sqrt 2 \).

Trong tam giác vuông SAO ta có :

\(SO = \sqrt {S{A^2} + A{O^2}} {\rm{\;}} = \sqrt {12{a^2} + 2{a^2}} {\rm{\;}} = a\sqrt {14} \).

\( \Rightarrow \tan \widehat {SOA} = \frac{{SA}}{{AO}} = \frac{{a\sqrt {14} }}{{a\sqrt 2 }} = \sqrt 7 {\rm{\;}} \Rightarrow \widehat {SOA} \approx 69^\circ 18'\).

Vậy \(\widehat {\left( {\left( {SBD} \right);\left( {ABCD} \right)} \right)} \approx 69^\circ 18'\).

d) Trong \(\left( {SAO} \right)\) kẻ \(AH \bot SO{\mkern 1mu} {\mkern 1mu} \left( {H \in SO} \right)\).

Ta có \(BD \bot \left( {SAC} \right) \Rightarrow BD \bot AH\).

\(\left\{ {\begin{array}{*{20}{l}}{AH \bot BD}\\{AH \bot SO}\end{array}} \right. \Rightarrow AH \bot \left( {SBD} \right) \Rightarrow d\left( {A;\left( {SBD} \right)} \right) = AH\).

Áp dụng hệ thức lượng trong tam giác vuông SOA ta có:

\(AH = \frac{{SA.AO}}{{\sqrt {S{A^2} + A{O^2}} }} = \frac{{2a\sqrt 3 .a\sqrt 2 }}{{\sqrt {12{a^2} + 2{a^2}} }} = \frac{{2\sqrt {21} a}}{7}\).

Vậy \(d\left( {A;\left( {SBD} \right)} \right) = \frac{{2\sqrt {21} a}}{7}\).

Trong \(\left( {SAB} \right)\), gọi \(I = AG \cap SB\) ta có: \(AG \cap \left( {SBD} \right) = I\).

\( \Rightarrow \frac{{d\left( {G;\left( {SBD} \right)} \right)}}{{d\left( {A;\left( {SBD} \right)} \right)}} = \frac{{GI}}{{AI}} = \frac{1}{3} \Rightarrow d\left( {G;\left( {SBD} \right)} \right) = \frac{1}{3}d\left( {A;\left( {SBD} \right)} \right) = \frac{{2\sqrt {21} a}}{{21}}\).