如图,在三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(CC_{1}⊥\)平面\(ABC\),\(D\),\(E\),\(F\),\(G\)分别为\(AA_{1}\),\(AC\),\(A_{1}C_{1}\),\(BB_{1}\)的中点,\(AB=BC= \sqrt {5}\),\(AC=AA_{1}=2\).
\((\)Ⅰ\()\)求证:\(AC⊥\)平面\(BEF\);
\((\)Ⅱ\()\)求二面角\(B-CD-C_{1}\)的余弦值;
\((\)Ⅲ\()\)证明:直线\(FG\)与平面\(BCD\)相交.