= Solution
Put $k_1=k_3=k$ and $m_1=m_2=m$. The mass-normalized stiffness <matrix> is
$$
\frac1m\begin{pmatrix}k+k_2&-k_2\\-k_2&k+k_2\end{pmatrix}.
$$
Its symmetric <eigenvector> $(1,1)$ has <eigenvalue> $k/m$, while its antisymmetric <eigenvector> $(1,-1)$ has <eigenvalue> $(k+2k_2)/m$. Thus
$$
\boxed{\Omega_{\rm sym}=\sqrt{\frac{k}{m}},\qquad\Omega_{\rm anti}=\sqrt{\frac{k+2k_2}{m}}}.
$$
The symmetric <frequency> does not depend on the middle <spring constant> because that <spring> has no extension in an in-phase <normal mode>. The antisymmetric motion stretches the middle <spring> and has the higher <frequency> for $k_2>0$.
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