某位射手每次射击命中目标的概率均为$p$，其中$0\lt p\lt 1$．

$(1)$当$p=\dfrac{2}{3}$时，若该射手射击$N$次，命中目标的次数为${X}$

①求$E\left( X \right)$

②若$P\left( X=7 \right)\gt P\left( X=k \right)$，其中$0\le k\le N$，且$k\ne 7$，求$N$的值

$(2)$某次射击游戏规则如下：若单次未命中目标得$0$分，单次命中目标得$1$分，若连续击中目标，第一次命中得$1$分，后续连续命中均得两分，记该射手射击$4$次总得分为$Y$，若对于任意$p$都有$P\left( Y=i \right)=P\left( Y=j \right)(i\lt j)$成立，求所有满足上述条件的有序实数对$\left( i,j \right)$

$(1)$①$\\dfrac{2}{3}N$ ；②$N=10$

$(2)$$\\left( 2,3 \\right),\\left( 4,5 \\right)$

$(1)$①由题意可知$X\sim B\left( N,\dfrac{2}{3} \right),\therefore E\left( X \right)=\dfrac{2}{3}N$

②设$P\left( X=n \right)$最大，则$\begin{cases} \text{\rm {C}}_{N}^{n}{{\left( \dfrac{2}{3} \right)}^{n}}{{\left( \dfrac{1}{3} \right)}^{N-n}}\ge \text{\rm {C}}_{N}^{n-1}{{\left( \dfrac{2}{3} \right)}^{n-1}}{{\left( \dfrac{1}{3} \right)}^{N-n+1}} \\ \text{\rm {C}}_{N}^{n}{{\left( \dfrac{2}{3} \right)}^{n}}{{\left( \dfrac{1}{3} \right)}^{N-n}}\ge \text{\rm {C}}_{N}^{n+1}{{\left( \dfrac{2}{3} \right)}^{n+1}}{{\left( \dfrac{1}{3} \right)}^{N-n-1}} \\ \end{cases}$

解得$\dfrac{2N-1}{3}\le n\le \dfrac{2N+2}{3}$，当$n=7$时，$N=10$或$N=11$

经检验，$N=11$时，$P\left( X=7 \right)=P\left( X=8 \right)$，不符合题意，舍去

$N=10$时，$P\left( X=7 \right)$为最大值，

$\therefore N=10$

$(2)$由题意该选手得分为$0,1,2,3,4,5,7$

$P\left( Y=0 \right)={{\left( 1-p \right)}^{4}}$

$P\left( Y=1 \right)=\text{\rm {C}}_{4}^{1}{{p}^{1}}{{\left( 1-p \right)}^{3}}$

$P\left( Y=2 \right)=3{{p}^{2}}{{\left( 1-p \right)}^{2}}$

$P\left( Y=3 \right)=3{{p}^{2}}{{\left( 1-p \right)}^{2}}$

$P\left( Y=4 \right)=2{{p}^{3}}\left( 1-p \right)$

$P\left( Y=5 \right)=2{{p}^{3}}\left( 1-p \right)$

$P\left( Y=7 \right)={{p}^{4}}$

满足题意的实数对为$\left( 2,3 \right),\left( 4,5 \right)$$.$