已知\(A\),\(F\)分别是椭圆\(C:\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a > b > 0)\)的左顶点和右焦点,点\(P\)为椭圆\(C\)上一动点,当\(PF⊥x\)轴时,\(|AF|=2|PF|\).
\((1)\)求椭圆\(C\)的离心率;
\((2)\)若椭圆\(C\)上存在点\(Q\),使得四边形\(AOPQ\)是平行四边形\((\)点\(P\)在第一象限\()\),求直线\(AP\)与\(OQ\)的斜率之积;
\((3)\)记圆\(O:{{x}^{2}}+{{y}^{2}}=\dfrac{ab}{{{a}^{2}}+{{b}^{2}}}\)为椭圆\(C\)的“关联圆”\(.\)若\(b=\sqrt{3}\),过点\(P\)作椭圆\(C\)的“关联圆”的两条切线,切点为\(M\),\(N\),直线\(MN\)的横、纵截距分别为\(m\),\(n\),求证:\(\dfrac{3}{{{m}^{2}}}+\dfrac{4}{{{n}^{2}}}\)为定值.