= Solution
Expand the second stage by the <Taylor theorem>, writing $f=f(t_n,y_n)$ and $D_yf$ for the derivative in the state variable:
$$
k_2=f+\frac h2\bigl(f_t+(D_yf)f\bigr)+O(h^2).
$$
The update therefore agrees with the exact solution's <Taylor series>:
$$
y_{n+1}=y_n+hf+\frac{h^2}{2}\bigl(f_t+(D_yf)f\bigr)+O(h^3).
$$
Thus the one-step <local truncation error> is $O(h^3)$, and the method has order at least two under the usual smoothness and <Lipschitz continuity> assumptions.
For $f=\lambda y$, the two-stage <Runge-Kutta method> gives
$$
\boxed{R(z)=1+z+\frac{z^2}{2}.}
$$
On the real axis, $R(z)=\tfrac12(z+1)^2+\tfrac12\geq\tfrac12$, so $|R(z)|\leq1$ is equivalent to $(z+1)^2\leq1$. Hence
$$
\boxed{\mathcal S\cap\mathbb R=[-2,0].}
$$
The decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the <linear stability domain>, the method is \b[not A-stable]. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.
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