如图\(1\),在高为\(2\)的梯形\(ABCD\)中,\(AB/\!/CD\),\(AB=2\),\(CD=5\),过\(A\)、\(B\)分别作\(AE⊥CD\),\(BF⊥CD\),垂足分别为\(E\)、\(F.\)已知\(DE=1\),将梯形\(ABCD\)沿\(AE\)、\(BF\)同侧折起,得空间几何体\(ADE-BCF\),如图\(2\).
\((\)Ⅰ\()\)若\(AF⊥BD\),证明:\(DE⊥BE\);
\((\)Ⅱ\()\)若\(DE/\!/CF\),\(CD= \sqrt {3}\),在线段\(AB\)上是否存在点\(P\)使得\(CP\)与平面\(ACD\)所成角的正弦值为\( \dfrac { \sqrt {35}}{35}\)?并说明理由.