= Solution
Every convergent member is also <A-stable>. For $0<\alpha<2$, put $\zeta=e^{i\theta}$ and calculate its <boundary-locus test for multistep A-stability>:
$$
\frac{\rho(\zeta)}\zeta=-\alpha(1-\cos\theta)+i(2-\alpha)\sin\theta,
\qquad
\frac{\sigma(\zeta)}\zeta=(2-\alpha)\cos\theta+i\frac\alpha2\sin\theta.
$$
The denominator never vanishes on the unit circle in this parameter interval. It follows that
$$
\operatorname{Re}\frac{\rho(e^{i\theta})}{\sigma(e^{i\theta})}
=\frac{\alpha(2-\alpha)(1-\cos\theta)^2}
{2[(2-\alpha)^2\cos^2\theta+\alpha^2\sin^2\theta/4]}\geq0.
$$
Therefore no amplification root can cross the unit circle as $z=h\lambda$ moves in the open left half-plane. For small negative real $z$, the root near one moves to $1+z+O(z^2)$, inside the circle; the other begins at $\alpha-1$, also strictly inside. The leading coefficient $1-(1-\alpha/4)z$ never vanishes in the left half-plane. Continuity of polynomial roots and connectedness of that half-plane now keep both roots inside everywhere. On its imaginary boundary the same locus has zero real part only at $\zeta=1$, corresponding to $z=0$; that root is simple because $2-\alpha>0$.
At $\alpha=0$, the two interlaced subsequences follow the <trapezoidal rule> with step $2h$:
$$
\zeta^2=\frac{1+z}{1-z}.
$$
The right side has modulus at most one for $\operatorname{Re}z\leq0$. Its unit-circle roots are simple; the double zero at $z=-1$ lies strictly inside and is harmless. Thus the endpoint is also <A-stable>. Combining this with part 1, the requested range is \b[$0\leq\alpha<2$], including the <BDF2 method> at $\alpha=4/3$.
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