$ABCD$ là hình vuông nên $BD\bot AC$
$SH\bot (ABCD)\to BD\bot SH$
Vậy $BD\bot (SAC)$
Ta có $\sin(AC,(SBC))=\sin(CH,(SBC))=\dfrac{d(H,(SBC))}{HC}$
$HC=\dfrac{3}{4}AC=\dfrac{3}{4}.a\sqrt2=\dfrac{3a\sqrt2}{4}$
Kẻ $HI//AB$
$AB\bot BC\to BC\bot HI$
Mà $SH\bot(ABCD)\to SH\bot BC$
Suy ra $BC\bot(SHI)$
Kẻ $HK\bot SI$
$BC\bot(SHI)\to BC\bot HK$
Suy ra $HK\bot(SBC)$
$\to d(H,(SBC))=HK$
Theo Talet, $\dfrac{HI}{AB}=\dfrac{CH}{CA}=\dfrac{3}{4}$
$\to HI=\dfrac{3}{4}a$
$\dfrac{1}{SH^2}+\dfrac{1}{HI^2}=\dfrac{1}{HK^2}$
$\to HK=\dfrac{3a}{5}$
Vậy $\sin(AC,(SBC))=\dfrac{2\sqrt2}{5}$
