(1)由题意$SD\bot $平面$ABCD$,连接$BD$,可得$SD\bot BD$,$BD\subset $平面$ABCD$,
那么直线$SB$与平面$ABCD$所成角为$\angle DSB$,
$\because \triangle SDB$是直角三角形,$SD=\sqrt{3}$.底面$ABCD$是边长为$1$的正方形,
$\therefore DB=\sqrt{2}$,那么$SB=\sqrt{5}$
$\therefore \cos \angle DSB=\frac{SD}{BD}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5}$.
$(2)$过$A$作$SB$的平行线交$BE$于$M$,可得$M$为直线$BE$上一点,
证明$DM$∥平面$SBC$,如下:
$\because AM$∥$SB,AD$∥$BC$,
$AM\cap AD=A$,$SB\cap BC=B$
$SB$、$BC\subset $平面$SBC$,$AM$、$AD\subset $平面$ADM$
$\therefore $平面$ADM$∥平面$SBC$,
而$MD\subset $平面$ADM$
$\therefore MD$∥平面$SBC$.
又$\because SE=2EA$,$\therefore EM=\frac{1}{2}BE$
$\because $底面$ABCD$是边长为$1$的正方形,$SD\bot $平面$ABCD$,
$\therefore EA\bot AB$,
$AB=1$,$AE=\frac{2}{3}$,$\therefore BE=\frac{\sqrt{13}}{3}$,
那么$BM=\frac{3}{2}BE=\frac{\sqrt{13}}{2}$.