= Solution
Introduce $t_0=t$, $T_1=\epsilon t$, $T=\epsilon^2t$ and write $x=x_0+\epsilon x_1+\cdots$. The <method of multiple scales> gives $x_0=A(T_1,T)e^{it_0}+\overline A e^{-it_0}$. At order $\epsilon$, the resonant forcing is $-2\partial_{t_0}\partial_{T_1}x_0$, because multiplication by $\cos t_0$ generates only a constant and second harmonics. The <solvability condition in the method of multiple scales> therefore gives $A_{T_1}=0$. A nonresonant particular correction is
$$
x_1=\frac A6e^{2it_0}+\frac{\overline A}6e^{-2it_0}-\frac{A+\overline A}{2}.
$$
At order $\epsilon^2$, the forcing is $-2\partial_{t_0}\partial_Tx_0-fx_0-\cos t_0x_1$. Its $e^{it_0}$ coefficient is $-2iA_T+(1/6-f)A+\overline A/4$. Removing this <secular term> gives
$$
2iA_T=(1/6-f)A+\overline A/4.
$$
Put $A=(a-ib)/2$, so $x_0=a(T)\cos t+b(T)\sin t$. The <slow amplitudes for the second Mathieu instability tongue> satisfy
$$
a_T=\frac{f+1/12}{2}b,\qquad b_T=\frac{5/12-f}{2}a,\qquad a(0)=1,\ b(0)=0.
$$
The initial conditions here are at leading order; a homogeneous $O(\epsilon)$ correction makes the original data exact without changing $x_0$. For example its initial cosine coefficient is $2/3$ in the chosen $x_1$ gauge. The slow squared growth rate is $(f+1/12)(5/12-f)/4$. Thus the robust leading stable ranges are
$$
\boxed{f<-1/12\quad\text{or}\quad f>5/12.}
$$
Set $\nu=\tfrac12\sqrt{(f+1/12)(f-5/12)}$ in those ranges. The required explicit leading solution is
$$
\boxed{x(t;\epsilon)=\cos(\nu\epsilon^2t)\cos t+
\frac{5/12-f}{2\nu}\sin(\nu\epsilon^2t)\sin t+O(\epsilon).}
$$
The remainder is uniform for $0\le\epsilon^2t\le C$, with fixed $C$ and fixed $f$ away from a changing instability edge. Inside $-1/12<f<5/12$, put $\sigma=\tfrac12\sqrt{(f+1/12)(5/12-f)}$; replace the slow cosine and $\sin(\nu T)/\nu$ by $\cosh(\sigma T)$ and $\sinh(\sigma T)/\sigma$. The initial data excite this growing solution, so the leading solution is unstable.
At the leading endpoints the slow matrix is degenerate. Its continuous limits give $x_0=\cos t+(T/4)\sin t$ at $f=-1/12$, and $x_0=\cos t$ for these particular initial data at $f=5/12$; generic initial phases at the latter endpoint have slow algebraic growth. These are leading, not exact, instability edges. To specify the exact small-$\epsilon$ endpoint status, put $t=2s$, so the <Mathieu equation> has characteristic parameter $a_M=4+4f\epsilon^2$ and coupling $q_M=-2\epsilon$. In the orthonormal cosine basis $1,\cos2s,\cos4s,\cos6s$, the matrix has diagonal $0,4,16,36$ and adjacent couplings $\sqrt2q_M,q_M,q_M$; in the sine basis $\sin2s,\sin4s,\sin6s$, the diagonal is $4,16,36$ and adjacent couplings are $q_M,q_M$. Expanding the roots near $4$ of their determinants yields
$$
a_2=4+\frac5{12}q_M^2-\frac{763}{13824}q_M^4+O(q_M^6),\quad
b_2=4-\frac1{12}q_M^2+\frac5{13824}q_M^4+O(q_M^6).
$$
These truncations determine the displayed orders because reaching the next omitted mode and returning requires at least six coupling steps. The <fourth-order edges of the second Mathieu instability tongue> are therefore
$$
f_-=-\frac1{12}+\frac5{3456}\epsilon^2+O(\epsilon^4),\qquad
f_+=\frac5{12}-\frac{763}{3456}\epsilon^2+O(\epsilon^4).
$$
<Floquet theory> gives the unstable gap $f_-<f<f_+$ near this resonance. In particular, both fixed values $-1/12$ and $5/12$ lie outside the exact gap for sufficiently small nonzero $\epsilon$. Their fixed-$\epsilon$ solutions are bounded, although the leading slow matrix alone does not show this and the amplification bound is not uniform as $\epsilon\to0$.
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