Bài 1:

Từ $A$ kẻ $AK\perp BD$

Ta có:

$\begin{cases}(SAB)\perp (ABCD)\\(SAD)\perp (ABCD)\\(SAB)\cap (SAD)= SA\end{cases}$

$\Rightarrow SA\perp (ABCD)$

$\Rightarrow SA\perp BD$

Khi đó:

$\begin{cases}AH\perp BD\quad \text{(cách dựng)}\\SA\perp BD\quad (cmt)\end{cases}$

$\Rightarrow BD\perp (SAH)$

Trong $mp(SAH)$ kẻ $AK\perp SH$

$\Rightarrow BD\perp AK$

$\Rightarrow AK\perp (SBD)\qquad (1)$

Mặt khác:

$\begin{cases}AD\perp AB\quad (gt)\\SA\perp AD\quad (SA\perp (ABCD))\end{cases}$

$\Rightarrow AD\perp (SAB)\qquad (2)$

$\Rightarrow \widehat{((SAB);(SBD))}=\widehat{KAD}= 45^\circ$

Do $AK\perp (SBD)$

nên $AK\perp KD$

$\Rightarrow \triangle KAD$ vuông tại $K$

Lại có: $\widehat{KAD}= 45^\circ$

$\Rightarrow \triangle KAD$ vuông cân tại $K$

$\Rightarrow AK = \dfrac{AD}{\sqrt2}= \dfrac{a}{\sqrt2}$

Ta có:

$AB;AD;SA$ đôi một vuông góc tại $A$

Do đó:

$\dfrac{1}{d^2(A;(SBD))}=\dfrac{1}{SA^2} + \dfrac{1}{AB^2} + \dfrac{1}{AD^2}$

$\Rightarrow \dfrac{1}{AK^2} = \dfrac{1}{SA^2} + \dfrac{1}{AB^2} + \dfrac{1}{AD^2}$

$\Rightarrow \dfrac{1}{SA^2} = \dfrac{1}{AK^2} - \dfrac{1}{AB^2} -\dfrac{1}{AD^2}$

$\Rightarrow \dfrac{1}{SA^2} = \dfrac{2}{a^2} - \dfrac{1}{4a^2} - \dfrac{1}{a^2}$

$\Rightarrow \dfrac{1}{SA^2}= \dfrac{4a^2}{3}$

$\Rightarrow SA = \dfrac{2a\sqrt3}{3}$

Khi đó:

$V_{S.ABC}= \dfrac12V_{S.ABCD}= \dfrac{1}{6}AB.AD.SA$

$\Rightarrow V_{S.ABC}=\dfrac16\cdot 2a\cdot a\cdot \dfrac{2a\sqrt3}{3}$

$\Rightarrow V_{S.ABC}=\dfrac{2a^3\sqrt3}{9}$

Bài 2:

Ta có: $SA\perp (ABC)$

$\Rightarrow \begin{cases}SA\perp AB\\SA\perp AC\end{cases}$

Do $SA,AB,AC$ đôi một vuông góc

Ta được:

$\dfrac{1}{d^2(A;(BCD))}=\dfrac{1}{SA^2} + \dfrac{1}{AB^2} +\dfrac{1}{AC^2}$

$\Rightarrow \dfrac{1}{SA^2}=\dfrac{1}{d^2(A;(BCD))} - \dfrac{1}{AB^2} - \dfrac{1}{AC^2}$

$\Rightarrow \dfrac{1}{SA^2}=\dfrac{9}{16a^2} - \dfrac{1}{4a^2} - \dfrac{1}{4a^2}$

$\Rightarrow \dfrac{1}{SA^2}=\dfrac{1}{16a^2}$

$\Rightarrow SA = 4a$

Khi đó:

$V_{S.ABC}=\dfrac16AB.AC.SA = \dfrac16\cdot 2a\cdot 2a\cdot 4a$

$\Rightarrow V_{S.ABC}= \dfrac{8a^3}{3}$