= Solution
{c}
The order-$\epsilon^2$ equation is
$$
(\partial_t^2+1)y_2=-2y_{0,tT}-a_2y_0-\cos t\,y_1.
$$
Using $\cos t\cos2t=(\cos3t+\cos t)/2$ and $\cos t\sin2t=(\sin3t+\sin t)/2$, its <resonance> coefficients are
$$
\cos t:\ -2B_T+(5/12-a_2)A,\qquad
\sin t:\ 2A_T-(a_2+1/12)B.
$$
<secular term> are removed by setting these to zero:
$$
A_T=\tfrac12(a_2+1/12)B,\qquad
B_T=\tfrac12(5/12-a_2)A.
$$
The <eigenvalues> of this two-by-two system satisfy
$$
\lambda^2=\frac14(a_2+1/12)(5/12-a_2).
$$
They are real with opposite signs for $-1/12<a_2<5/12$. Generic initial amplitudes then contain an exponentially growing slow-time component. At either leading endpoint the <matrix> is nonzero <nilpotent matrix>, so suitable initial amplitudes give linear slow-time growth; higher-order detuning decides the precise corrected boundary. Outside this interval the leading <eigenvalues> are purely imaginary and the slow amplitudes remain oscillatory.
Substitution into $a=1+a_2\epsilon^2+\cdots$ gives the requested leading instability interval
$$
\boxed{1-\frac{\epsilon^2}{12}+O(\epsilon^3)
\leq a\leq1+\frac{5\epsilon^2}{12}+O(\epsilon^3).}
$$
“Can become unbounded” permits special initial amplitudes without growth. The calculation identifies the <second instability tongue of a weak Mathieu oscillator> and its <slow amplitudes for the second Mathieu instability tongue>; it is not an assertion that every solution grows throughout the closed interval.
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