如图所示,在三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(AC=BC\),\(AA_{1}⊥\)平面\(ABC\),点\(D\),\(D_{1}\)分别是\(AB\),\(A_{1}B_{1}\)的中点.
\((1)\)求证:平面\(AC_{1}D_{1}/\!/\)平面\(CDB_{1}\);
\((2)\)求证:平面\(CDB_{1}⊥\)平面\(ABB_{1}A_{1}\);
\((3)\)若\(AC⊥BC\),\(AC=AA_{1}\),求异面直线\(AC_{1}\)与\(A_{1}B\)所成的角.