如图，在三棱柱$ABC-A_{1}B_{1}C_{1}$中，$BB_{1}\bot $平面$ABC$，$AB\bot BC$，$AA_{1}=AB=BC=2$.

$(1)$求证：$BC_{1}\bot $平面$A_{1}B_{1}C$；
$(2)$求二面角$B_{1}-A_{1}C-C_{1}$的余弦值：
$(3)$点$M$在线段$B_{1}C$上，且$\dfrac{{B_1}M}{{B_1}C}=\dfrac{1}{3}$，点$N$在线段$A_{1}B$上，若$MN//$平面$A_{1}ACC_{1}$，求$\dfrac{{A}_{1}N}{{A}_{1}B}$的值.

$(1)$证明见解析；

$(2)$$\\dfrac{1}{2}$；

$(3)$$\\dfrac{2}{3}$

$(1)$证明：在三棱柱$ABC-A_{1}B_{1}C_{1}$中，$BB_{1}\bot $平面$ABC$，$AB\bot BC$，
因为平面$ABC//$平面$A_{1}B_{1}C_{1}$，
所以$BB_{1}\bot $平面$A_{1}B_{1}C_{1}$，
又$A_{1}B_{1}\subset $平面$A_{1}B_{1}C_{1}$，$B_{1}C_{1}\subset $平面$A_{1}B_{1}C_{1}$，
所以$BB_{1}\bot A_{1}B_{1}$，且$BB_{1}\bot B_{1}C_{1}$，
又$A_{1}B_{1}\bot B_{1}C_{1}$，
因为$BB_{1}\cap B_{1}C_{1}=B_{1}$，
所以$A_{1}B_{1}\bot $平面$BBC_{1}B_{1}$，
因为$BC_{1}\subset $平面$BBC_{1}B_{1}$，
所以$A_{1}B_{1}\bot BC_{1}$，
由$BB_{1}=AA_{1}=BC$，则侧面$BB_{1}C_{1}C$为正方形，
所以$BC_{1}\bot B_{1}C$，
因为$A_{1}B_{1}\cap B_{1}C=B_{1}$，$A_{1}B_{1}\subset $平面$A_{1}B_{1}C$，$B_{1}C\subset $平面$A_{1}B_{1}C$，
所以$BC_{1}\bot $平面$A_{1}B_{1}C$.
$(2)$以$B$为原点，$BC$为$x$轴，$BA$为$y$轴，$BB_{1}$为$z$轴，建立空间直角坐标系，如图，
，
$A\left(0,2,0\right)$，$C\left(2,0,0\right)$，$C_{1}(2,0,2)$，$B\left(0,0,0\right)$，$B_{1}(0,0,2)$，$A_{1}(0,2,2)$，
所以$\overrightarrow{CA}=(-2,2,0)$，$\overrightarrow{C{C}_{1}}=(0,0,2)$，$\overrightarrow{{B}_{1}{A}_{1}}=(0,2,0)$，$\overrightarrow{{B}_{1}C}=(2,0,-2)$，
设平面$ACC_{1}A_{1}$的法向量$\boldsymbol{n}=({x}_{1},{y}_{1},{z}_{1})$，
则$\left\{\begin{array}{l}{\boldsymbol{n}\cdot \overrightarrow{CA}=({x}_{1},{y}_{1},{z}_{1})\cdot (-2,2,0)=-2{x}_{1}+2{y}_{1}=0}\\{\boldsymbol{n}\cdot \overrightarrow{C{C}_{1}}=({x}_{1},{y}_{1},{z}_{1})\cdot (0,0,2)=2{z}_{1}=0}\end{array}\right.$，
取$x_{1}=1$，则$y_{1}=1$，$z_{1}=0$，
所以$\boldsymbol{n}=\left(1,1,0\right)$，
设平面$B_{1}A_{1}C$的法向量$\boldsymbol{m}=({x}_{2},{y}_{2},{z}_{2})$，
则$\left\{\begin{array}{l}{\boldsymbol{m}\cdot \overrightarrow{{B}_{1}C}=({x}_{2},{y}_{2},{z}_{2})\cdot (2,0,-2)=2{x}_{2}-2{z}_{2}=0}\\{\boldsymbol{m}\cdot \overrightarrow{{B}_{1}{A}_{1}}=({x}_{2},{y}_{2},{z}_{2})\cdot (0,2,0)=2{y}_{2}=0}\end{array}\right.$，
取$x_{2}=1$，则$y_{2}=0$，$z=1$，
所以$\boldsymbol{m}=(1,0,1)$，
设二面角$B_{1}-A_{1}C-C_{1}$的平面角为$\theta $，
则$|\cos\theta |=|\cos⟨\boldsymbol{m},\boldsymbol{n}⟩|=\dfrac{|\boldsymbol{m}\cdot \boldsymbol{n}|}{|\boldsymbol{m}||\boldsymbol{n}|}=\dfrac{1}{\sqrt{2}×\sqrt{2}}=\dfrac{1}{2}$，
因为二面角$B_{1}-A_{1}C-C_{1}$为锐二面角，
所以二面角$B_{1}-A_{1}C-C_{1}$的余弦值为$\dfrac{1}{2}$.
$(3)$点$M$在线段$B_{1}C$上，且$\dfrac{{B_1}M}{{B_1}C}=\dfrac{1}{3}$，点$N$在线段$A_{1}B$上，
设$M\left(a,b,c\right)$，$N\left(x,y,z\right)$，
设$\dfrac{{A_1}N}{{A_1}B}=\lambda $，则$\overrightarrow{{B}_{1}C}=3\overrightarrow{{B}_{1}M}$，$\overrightarrow{{A}_{1}N}=\lambda \overrightarrow{{A}_{1}B}$，且$0\leqslant \lambda \leqslant 1$，
且$\overrightarrow{{A}_{1}B}=(0,-2,-2)$，
即$\left(2,0,-2\right)=3\left(a,b,c-2\right)$，$\left(x,y-2,z-2\right)=\lambda \left(0,-2,-2\right)$，
解得$M\left( \dfrac{2}{3},0,\dfrac{4}{3}\right)$，$N\left(0,2-2\lambda ,2-2\lambda \right)$，
$\overrightarrow{MN}=\left( -\dfrac{2}{3},2-2\lambda,\dfrac{2}{3}-2\lambda\right)$，
因为$MN$∥平面$ACC_{1}A_{1}$，且$\boldsymbol{n}=(1,1,0)$，
所以$\boldsymbol{n}\cdot \overrightarrow{MN}=-\dfrac{2}{3}+2-2\lambda =0$，解得$\lambda =\dfrac{2}{3}$.
所以$\dfrac{{A}_{1}N}{{A}_{1}B}$的值为$\dfrac{2}{3}$.