= Solution
For fixed $L$, expansion of the matrix in part (b) gives
$$
(M_\varepsilon^L)^TM_\varepsilon^L
=
\begin{pmatrix}
(1+O(\varepsilon^4))I&O(\varepsilon^3)I\\
O(\varepsilon^3)I&(1+O(\varepsilon^4))I
\end{pmatrix}.
$$
Hence its largest eigenvalue is $1+O(\varepsilon^3)$. With $z=(x(0),p(0))$ and $z'=M_\varepsilon^Lz$,
$$
H(z')=\frac12\|z'\|^2
\leq(1+C_0\varepsilon^3)H(z).
$$
For each fixed starting state, or uniformly on any bounded set of starting states, this implies
$$
H(z')-H(z)\leq C_1\varepsilon^3.
$$
Using $e^{-u}\geq1-u$ for $u\geq0$ in the acceptance formula gives
$$
\boxed{1-C\varepsilon^3\leq\alpha\leq1}
$$
for sufficiently small $\varepsilon$. The constant necessarily depends on a bound for $\|z\|$: a uniform pointwise constant over all of $\mathbb R^{2d}$ would be impossible because the energy error is quadratic in the starting state.
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