 Solution

For the biquadratic extension $L=K(\sqrt{a},\sqrt{b})$, the assumptions say that the square classes of $a$ and $b$ are independent. In particular $\sqrt{b}\notin K(\sqrt{a})$: writing $\sqrt{b}=u+v\sqrt{a}$ and comparing the coefficient of $\sqrt{a}$ after squaring would force either $b$ or $ab$ to be a square in $K$. Hence

The field is the splitting field of $(X^2-a)(X^2-b)$ and the characteristic is not two, so it is Galois. Its four automorphisms independently choose the signs of $\sqrt{a}$ and $\sqrt{b}$, givingBy the Galois correspondence, the intermediate-field lattice isEach of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.