= Solution
For the linear test equation $f(y)=\lambda y$ put $z=h\lambda$. The second stage solves $(1-az)k_2=\lambda y[1+(1-a)z]$, and substitution into the update gives the <stability function>
$$
\boxed{R(z)=\frac{1+(1-a)z+(1/2-a)z^2}{1-az}.}
$$
<A-stability> requires no <poles> and $|R(z)|\leq1$ throughout the closed left half-plane. If $a\ne1/2$, the quadratic numerator gives unbounded magnitude as $z\to-\infty$; when $a=0$ the denominator is constant and the same conclusion holds. Thus $a=1/2$ is necessary. At this value $R(z)=(1+z/2)/(1-z/2)$, whose only <pole> is at positive $z=2$. Moreover
$$
|1-z/2|^2-|1+z/2|^2=-2\operatorname{Re}z\geq0
$$
on the left half-plane. Hence \b[the method is A-stable exactly for $\boxed{a=1/2}$]. Its limit at negative infinity is $-1$, so it is not <L-stable>.
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