= Solution
The time-independent <Schrödinger equation> is
$$
-\frac{\hbar^2}{2m}\psi''(x)+V(x)\psi(x)=E\psi(x).
$$
Integrating it over $(a-\varepsilon,a+\varepsilon)$ and letting $\varepsilon\to0$ shows that $\psi'$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in $\psi'$ and hence a <derivative> of a delta in $\psi''$, so $\psi$ is also continuous.
For a <bound state> $-V_0<E<0$, define
$$
\eta^2=-\frac{2mE}{\hbar^2},
\qquad
k^2=\frac{2m(E+V_0)}{\hbar^2}.
$$
Then the proposed even <wavefunction> solves the equation away from the interfaces, and
$$
\boxed{k^2+\eta^2=\frac{2mV_0}{\hbar^2}}.
$$
Continuity at $x=a$ gives $B\cos(ka)=Ae^{-\eta a}$, while continuity of the <derivative> gives $-Bk\sin(ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relation
$$
\boxed{\eta=k\tan(ka)}.
$$
\b[Thus the lowest even state of the <finite square well> is the first-quadrant intersection, with $0<ka<\pi/2$, of the circle $k^2+\eta^2=2mV_0/\hbar^2$ and the curve $\eta=k\tan(ka)$.]
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