If $e_1, e_2$ are the eccentricities of the hyperbola $2x^2-2y^2=1$and the ellipse $x^2+2y^2=2$ respectively. Then

Equation of the hyperbola is $\frac{x^2}{1/2}-\frac{y^2}{1/2}=1$ 

So $e_1^2=\frac{a^2+b^2}{a^2}$ where $a^2=b^2=\frac{1}{2}$ $\Rightarrow e_1=\sqrt{2}$

(Note: It is a rectangular hyperbola)

Equation of the ellipse is $\frac{x^2}{2}+\frac{y^2}{1}=1$

$\text{ So }\begin{array}{l}{e}_2^2=\frac{a^2-b^2}{a^2}=\frac{2-1}{2}=\frac{1}{2}\\ \Rightarrow e_2=1/\sqrt{2}\text{ hence }{e}_1{e}_2=1.\end{array}$