$\left(1\right)\because$反比例函数$y=\dfrac{2}{x}$的图象和一次函数的图象交于$A$、$B$两点，点$A$的横坐标是$-1$，点$B$的纵坐标是$-1$，

$\therefore A\left(-1,-2\right)$，$B\left(-2,-1\right)$，

设一次函数的表达式为$y=kx+b$，

$\because$经过$A\left(-1,-2\right)$，$B\left(-2,-1\right)$点，

$\therefore \left\{\begin{array}{l}-k+b=-2\\-2k+b=-1\end{array}\right.$，解得$\left\{\begin{array}{l}k=-1\\b=-3\end{array}\right.$，

$\therefore$这个一次函数的表达式为$y=-x-3$；

$\left(2\right)\because$点$P\left(m,n\right)$与点$Q$关于$x$轴对称，

$\therefore Q\left(m,-n\right)$，

$\because$点$P\left(m,n\right)$在反比例函数图象上，

$\therefore mn=2$，

$\because$点$Q$恰好落在一次函数的图象上，

$\therefore n=m+3$，

$\therefore m\left(m+3\right)=2$，

$\therefore m^{2}+3m=2,$

$\therefore m^{2}+n^{2}=m^{2}+\left(m+3\right)^{2}=2m^{2}+6m+9=2\left(m^{2}+3m\right)+9=2\times 2+9=13$；

（3）如图，过$M$作$MG\bot x$轴于$G$，过$N$作$NH\bot x$轴于$H$，

$\because M(x_{1}$，$y_{1})$，$N(x_{2}$，$y_{2})$是反比例函数$y=\dfrac{2}{x}$在第一象限图象上的两点，

$\therefore S_{\triangle MOG}=S_{\triangle NOH}=\dfrac{1}{2}|k|=1$，

$\because x_{2}-x_{1}=2$，$y_{1}+y_{2}=3$，

$\therefore S_{\triangle MON}=S_{梯形MNHG}+S_{\triangle MOG}-S_{\triangle NOH}=S_{梯形MNHG}=\dfrac{1}{2}\left(y_{1}+y_{2}\right)\left(x_{2}-x_{1}\right)=\dfrac{1}{2}\times 3\times 2=3$.

$\left(1\right)\because$反比例函数$y=\dfrac{2}{x}$的图象和一次函数的图象交于$A$、$B$两点，点$A$的横坐标是$-1$，点$B$的纵坐标是$-1$，

$\therefore A\left(-1,-2\right)$，$B\left(-2,-1\right)$，

设一次函数的表达式为$y=kx+b$，

$\because$经过$A\left(-1,-2\right)$，$B\left(-2,-1\right)$点，

$\therefore \left\{\begin{array}{l}-k+b=-2\\-2k+b=-1\end{array}\right.$，解得$\left\{\begin{array}{l}k=-1\\b=-3\end{array}\right.$，

$\therefore$这个一次函数的表达式为$y=-x-3$；

$\left(2\right)\because$点$P\left(m,n\right)$与点$Q$关于$x$轴对称，

$\therefore Q\left(m,-n\right)$，

$\because$点$P\left(m,n\right)$在反比例函数图象上，

$\therefore mn=2$，

$\because$点$Q$恰好落在一次函数的图象上，

$\therefore n=m+3$，

$\therefore m\left(m+3\right)=2$，

$\therefore m^{2}+3m=2,$

$\therefore m^{2}+n^{2}=m^{2}+\left(m+3\right)^{2}=2m^{2}+6m+9=2\left(m^{2}+3m\right)+9=2\times 2+9=13$；

（3）如图，过$M$作$MG\bot x$轴于$G$，过$N$作$NH\bot x$轴于$H$，

$\because M(x_{1}$，$y_{1})$，$N(x_{2}$，$y_{2})$是反比例函数$y=\dfrac{2}{x}$在第一象限图象上的两点，

$\therefore S_{\triangle MOG}=S_{\triangle NOH}=\dfrac{1}{2}|k|=1$，

$\because x_{2}-x_{1}=2$，$y_{1}+y_{2}=3$，

$\therefore S_{\triangle MON}=S_{梯形MNHG}+S_{\triangle MOG}-S_{\triangle NOH}=S_{梯形MNHG}=\dfrac{1}{2}\left(y_{1}+y_{2}\right)\left(x_{2}-x_{1}\right)=\dfrac{1}{2}\times 3\times 2=3$.