= Solution
Use the unitary change of basis from part (b). It reduces the <Feynman-Kitaev Hamiltonian> to $A+B$, where
$$
A=Q\otimes|0\rangle\langle0|,\qquad B=I\otimes E.
$$
The positive <eigenvalues> of $Q$ are positive integers, since its commuting ancilla projectors act on different qubits. The positive <spectral gap> of $B$ is $1-\cos(\pi/(T+1))\geq2/(T+1)^2$. Thus both positive spectra are bounded below by $\mu\geq2/(T+1)^2$, for $T\geq1$.
Let $|u\rangle=(T+1)^{-1/2}\sum_t|t\rangle$, and split the work space into $G=\ker Q$ and $G^\perp$. The common <ground space> is $K=G\otimes|u\rangle$. A <unit vector> in $\ker B$ orthogonal to $K$ has the form $|z\rangle|u\rangle$, where $z\in G^\perp$. Its <orthogonal projection> onto $\ker A$ simply removes its time-zero component, so the projected norm is $\sqrt{T/(T+1)}$. Consequently the <smallest angle between two subspaces>, after removing their common intersection, satisfies
$$
\cos\vartheta=\sqrt{\frac T{T+1}},\qquad \sin^2\vartheta=\frac1{T+1}.
$$
The <Kitaev geometrical lemma> now gives
$$
\Delta(H)\geq2\mu\sin^2(\vartheta/2)=\mu(1-\cos\vartheta)\geq\frac\mu{2(T+1)}\geq\frac1{(T+1)^3}.
$$
Here $1-\sqrt{1-x}\geq x/2$ supplies the penultimate step. Hence
$$
\boxed{\Delta(H)=\Omega(T^{-3}).}
$$
If there are no input constraints, $Q=0$ and the propagation gap is already $\Omega(T^{-2})$, which is stronger.
The printed geometric-lemma notation needs a correction: the maximum overlap defines $\cos\vartheta$, not $\vartheta$, and is taken over normalized vectors in the two kernels, with the common <ground space> removed. The <ground space> restriction is essential when many <quantum witnesses> are allowed.
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