$\dfrac{x^2}4+\dfrac{y^2}2=1$．

本题考查椭圆的基本量．由题，知 $b=\sqrt2$，$e=\dfrac{c}{a}=\dfrac{\sqrt2}{2}$，再结合 $a^2=b^2+c^2$，解得 $a=2,b=\sqrt2$，因此椭圆 $E$ 的方程为 $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$．

点 $G\left(-\dfrac 94,0\right)$ 在以 $AB$ 为直径的圆外．

本题考查直线与圆锥曲线的位置关系，可将点与圆的位置关系转化为对应角度的锐直钝问题．解法一：
设 $A\left(x_1,y_1\right)$，$B\left(x_2,y_2\right)$，$AB$ 的中点为 $H\left(x_0,y_0\right)$．
联立直线与椭圆方程，得\[\begin{cases}
x=my-1,\\
\dfrac{x^2}4+\dfrac{y^2}2=1,
\end{cases}\]整理得\[\left(m^2+2\right)y^2-2my-3=0,\]所以\[\begin{cases}y_1+y_2=\dfrac{2m}{m^2+2},\\ y_1y_2=-\dfrac 3{m^2+2},\end{cases}\]从而 $y_0=\dfrac m{m^2+2}$．所以\[\begin{split}\left|GH\right|^2&=\left(x_0+\dfrac 94\right)^2+y_0^2\\&=\left(my_0+\dfrac 54\right)^2+y_0^2\\&=\left(m^2+1\right)y_0^2+\dfrac 52my_0+\dfrac{25}{16}.\end{split}\]\[\begin{split}\dfrac{\left|AB\right|^2}4&=\dfrac{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}4
\\&=\dfrac{\left(1+m^2\right)\left(y_1-y_2\right)^2}4
\\&=\dfrac{\left(1+m^2\right)\left[\left(y_1+y_2\right)^2-4y_1y_2\right]}4
\\&=\left(1+m^2\right)\left(y_0^2-y_1y_2\right),
\end{split}\]所以，\[\begin{split}\left|GH\right|^2-\dfrac{\left|AB\right|^2}4&=\dfrac 52my_0+\left(1+m^2\right)y_1y_2+\dfrac{25}{16}\\&=\dfrac{5m^2}{2\left(m^2+2\right)}-\dfrac{3\left(1+m^2\right)}{m^2+2}+\dfrac{25}{16}
\\&=\dfrac{17m^2+2}{16\left(m^2+2\right)}>0,\end{split}\]所以 $\left|GH\right|>\dfrac{\left|AB\right|}{2}$，
故点 $G\left(-\dfrac 94,0\right)$ 在以 $AB$ 为直径的圆外．
解法二：
设点 $A\left(x_1,y_1\right)$，$B\left(x_2,y_2\right)$，则 $\overrightarrow{GA}=\left(x_1+\dfrac 94,y_1\right)$，$\overrightarrow{GB}=\left(x_2+\dfrac 94,y_2\right)$．
联立直线与椭圆方程，得\[\begin{cases}
x=my-1,\\
\dfrac{x^2}4+\dfrac{y^2}2=1,
\end{cases}\]整理得\[\left(m^2+2\right)y^2-2my-3=0,\]所以\[\begin{cases}y_1+y_2=\dfrac{2m}{m^2+2},\\ y_1y_2=-\dfrac 3{m^2+2},\end{cases}\]因此，\[\begin{split}\overrightarrow{GA}\cdot\overrightarrow{GB}&\overset{\left[a\right]}=\left(x_1+\dfrac 94\right)\left(x_2+\dfrac 94\right)+y_1y_2
\\&=\left(my_1+\dfrac 54\right)\left(my_2+\dfrac 54\right)+y_1y_2
\\&=\left(m^2+1\right)y_1y_2+\dfrac 54m\left(y_1+y_2\right)+\dfrac{25}{16}
\\&=\dfrac{-3\left(m^2+1\right)}{m^2+2}+\dfrac{\dfrac 52m^2}{m^2+2}+\dfrac{25}{16}
\\&=\dfrac{17m^2+2}{16\left(m^2+2\right)}>0 ,
\end{split}\]（推导中用到[a]）
所以 $\cos\left\langle \overrightarrow{GA},\overrightarrow{GB}\right\rangle>0$．
又 $\overrightarrow{GA}$，$\overrightarrow{GB}$ 不共线，所以 $\angle AGB$ 为锐角．
故点 $G\left(-\dfrac 94,0\right)$ 在以 $AB$ 为直径的圆外．