a,
Gọi $I$ là trung điểm $CD$
$HI//AD\to CD\bot HI$ vì $AD\bot CD$
Kẻ $HK\bot SI$
Có $CD\bot HI, CD\bot SH$ nên $CD\bot (SHI)$
$\to CD\bot HK$
$\to HK\bot (SCD)$
$\to d(H,(SCD))=HK$
Có $HI=BC=2a$
$\dfrac{1}{SH^2}+\dfrac{1}{HI^2}=\dfrac{1}{HK^2}$
$\to HK=\dfrac{2a\sqrt{21}}{7}$
Vậy $d(H,(SCD))=\dfrac{2a\sqrt{21}}{7}$
b,
Gọi $E=HC\cap MD$
Theo tính chất hình vuông, ta có $MD\bot HE$
Kẻ $HF\bot SE$
Có $MD\bot HE, MD\bot SH$ nên $MD\bot (SHE)$
$\to MD\bot HF$
Suy ra $HF\bot (SMD)$
$\to d(H,(SMD))=HF$
$BH=MC=\dfrac{BC}{2}=a$
$\to HC=\sqrt{a^2+(2a)^2}=a\sqrt5$
$\Delta MEC\backsim\Delta HBC(g.g)$ nên ta có:
$\dfrac{EC}{MC}=\dfrac{BC}{HC}$
$\to EC=\dfrac{a.2a}{a\sqrt5}=\dfrac{2a}{\sqrt5}$
$\to HE=HC-EC=\dfrac{3a}{\sqrt5}$
$\dfrac{1}{SH^2}+\dfrac{1}{HE^2}=\dfrac{1}{HF^2}$
$\to HF=\dfrac{3a\sqrt2}{4}$
Vậy $d(H,(SMD))=\dfrac{3a\sqrt2}{4}$
