$解：?(2)?設(shè)?△ABC?的姊妹三角形為?△DEF，?且?DE=DF，?如圖$
$∵在?△ABC?中，?AB=AC，??∠A=30°，??BC=\sqrt 6-\sqrt 2?$
$∴?∠B=∠C=75°?$
$過(guò)點(diǎn)?B?作?BG⊥AC，?垂足為點(diǎn)?G，?設(shè)?BG=x，?則?AB=AC=2x，??AG=\sqrt 3x?$
$∴?CG=AC-AG=2x-\sqrt 3x=(2-\sqrt 3)x?$
$在?Rt△BGC?中，?BG^2+CG^2=BC^2?$
$∴?x^2+(2-\sqrt 3)^2x^2=(\sqrt 6-\sqrt 2)^2?$
$∴?x=1?$
$∴?AB=AC=2?$
$?①∠D=∠ABC=75°，??DE=DF=BC=\sqrt 6-\sqrt 2?$
$②當(dāng)?∠E=∠A=30°?時(shí)，?∠EDF=120°，??EF=AB=2，?如圖$
$過(guò)點(diǎn)?D?作?DH⊥EF，?垂足為點(diǎn)?H?$
$∵?DE=DF?$
$∴?EH=\frac 12EF=1?$
$∴?ED=\frac {EH}{cos 30°}=\frac {2\sqrt 3}3?$
$∴?△ABC?的姊妹三角形的頂角為?75°?時(shí)，腰長(zhǎng)為?\sqrt 6-\sqrt 2?$
$頂角為?120°?時(shí)，腰長(zhǎng)為?\frac {2\sqrt 3}3?$
