= Solution
Choose mass-weighted coordinates $Q_A$, $\mathbf Q_E=(Q_{E1},Q_{E2})$, and two triplets $\mathbf Q_3,\mathbf Q_4$ for the two $T_2$ copies. About a stationary equilibrium geometry, the allowed harmonic potential is
$$
V^{(2)}=\frac12k_AQ_A^2+\frac12k_E|\mathbf Q_E|^2
+\frac12k_{33}|\mathbf Q_3|^2+k_{34}\mathbf Q_3\cdot\mathbf Q_4
+\frac12k_{44}|\mathbf Q_4|^2.
$$
The <kinetic energy> is one-half the sum of the squared conjugate momenta. Cross terms between inequivalent <irreducible representations> are forbidden, but the two equivalent $T_2$ triplets permit the displayed scalar-product term. This is <equivalent-mode harmonic coupling>. An orthogonal rotation between the two triplets diagonalizes their $2\times2$ force-constant matrix. In <molecular normal coordinates> the complete <Hamiltonian> becomes
$$
\boxed{H=E_0+\frac12(P_A^2+\omega_1^2Q_A^2)
+\frac12\sum_{j=1}^2(P_{Ej}^2+\omega_2^2Q_{Ej}^2)
+\frac12\sum_{j=1}^3(P_{3j}^2+\omega_3^2Q_{3j}^2+P_{4j}^2+\omega_4^2Q_{4j}^2)}.
$$
Symmetry enforces degeneracies within each multiplet; it does not require equal frequencies for the two distinct $T_2$ <normal modes>.
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