= Solution
The occupation formula from part (b) gives
$$
n_2=f_2^\dagger f_2=\frac{1+i\gamma_1^B\gamma_2^A}{2}.
$$
Consequently
$$
\boxed{H=t(2f_2^\dagger f_2-1)=t(2n_2-1)}.
$$
The independent <occupation numbers> $n_1,n_2$ are each zero or one. The <Hamiltonian> is independent of $n_1$, so its four states have
$$
\boxed{\begin{array}{c|c|c}
n_2&E&\text{degeneracy}\\\hline
0&-t&2\\
1&+t&2
\end{array}}.
$$
Each energy admits $n_1=0$ or $n_1=1$; these states differ by occupancy of the zero-energy mode. The ground energy is $-|t|$ for either sign of $t$. These degeneracies refer to the full two-mode <fermionic Fock space>. If one separately restricts to a fixed global <fermion parity>, only one state at each energy remains in that parity sector.
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