= 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
$$
[L:K]=[K(\sqrt a,\sqrt b):K(\sqrt a)]
\,[K(\sqrt a):K]=4.
$$
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 $\sqrt b$, giving
$$
\operatorname{Gal}(L/K)\cong C_2\times C_2.
$$
By the <Galois correspondence>, the intermediate-field lattice is
$$
\begin{array}{c}
L\\[2mm]
K(\sqrt a)\qquad K(\sqrt b)\qquad K(\sqrt{ab})\\[2mm]
K.
\end{array}
$$
Each of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.
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