Let the equation of ellipse $$\mathrm{E}_2$$ is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$.

Now, it is given that $$\mathrm{E}_2$$ passes through $$(0,4)$$

$$\begin{aligned} \Rightarrow \quad \frac{0^2}{a^2}+\frac{4^2}{b^2} & =1 \\ b^2 & =16 \end{aligned}$$

Also, $$\mathrm{E}_2$$ passes through $$( \pm 3, \pm 2)$$

$$\begin{aligned} & \Rightarrow \quad \frac{3^2}{a^2}+\frac{2^2}{b^2}=1 \\ & \Rightarrow \quad \frac{9}{a^2}+\frac{4}{16}=1 \\ & \Rightarrow \quad \frac{9}{a^2}=\frac{3}{4} \\ & \Rightarrow \quad a^2=12 \\ & \text { Now, } \quad b^2>a^2 \\ & \Rightarrow \quad b>a \\ & \therefore \quad a^2=b^2\left(1-e^2\right) \\ & \Rightarrow \quad 12=16\left(1-e^2\right) \\ & \Rightarrow \quad \frac{12}{16}=1-e^2 \\ & \Rightarrow \quad e^2=1-\frac{3}{4} \\ & \Rightarrow \quad e^2=\frac{1}{4} \\ & \Rightarrow \quad e=\frac{1}{2} \\ \end{aligned}$$