= Solution
For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this <implicit Runge-Kutta method> are
$$
\begin{pmatrix}
1-z/4&-z(1/4-a)\\
-z(1/4+a)&1-z/4
\end{pmatrix}
\begin{pmatrix}k_1\\k_2\end{pmatrix}
=\lambda y_n\begin{pmatrix}1\\1\end{pmatrix}.
$$
Substitution into the update gives the <stability function>
$$
R(z)=\frac{2+z+2a^2z^2}{2-z+2a^2z^2}.
$$
For $z=x+iy$, direct expansion gives
$$
\begin{aligned}
&|2-z+2a^2z^2|^2-|2+z+2a^2z^2|^2\\
&\hspace{35mm}=-8x(1+a^2|z|^2).
\end{aligned}
$$
This is nonnegative whenever $x\leq0$, and hence $|R(z)|\leq1$ throughout the closed left half-plane, provided the denominator has no zero there.
If $a=0$, the denominator vanishes only at $z=2$. If $a\ne0$, its zeros solve
$$
2a^2z^2-z+2=0.
$$
Real roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Therefore
$$
\boxed{a\in\mathbb R}
$$
is the complete set of <A-stable> parameters, as summarized by the <A-stability of a symmetric two-stage implicit Runge-Kutta family>.
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