对于数列\(\{a_{n}\}\),定义\(b_{n}(k)=a_{n}+a_{n+k}\),其中\(n\),\(k∈N\).
\((1)\)若\(b_{n}(2)-b_{n}(1)=1\),\(n∈N^{*}\),求\(b_{n}(4)-b_{n}(1)\)的值;
\((2)\)若\(a_{1}=2\),且对任意的\(n\),\(k∈N^{*}\),都有\(b_{n+1}(k)=2b_{n}(k)\).
\((ⅰ)\)求数列\(\{a_{n}\}\)的通项公式;
\((ⅱ)\)设\(k\)为给定的正整数,记集合\(A=\{b_{n}(k)|n∈N^{*}\}\),\(B=\{5b_{n}(k+2)|n∈N^{*}\}\),求证:\(A∩B=\varnothing \).