= Solution
With $q=2\lambda+1$, the reduced coefficient matrix and its symmetric part are
$$
A=\begin{pmatrix}-q^2/2&2\alpha\\-2\alpha&-(q^2+1)/2\end{pmatrix},
\qquad
A_s=\frac{A+A^T}{2}
=-\frac12\begin{pmatrix}q^2&0\\0&q^2+1\end{pmatrix}.
$$
Its diagonal entries are both strictly negative exactly when $q\ne0$, equivalently $\boxed{\lambda\ne-1/2}$. The coherent driving cancels from the symmetric part because it is a rotation generator.
For a <steady state> $s_*$ put $e=s-s_*$. Differentiating the squared Euclidean distance gives
$$
\frac d{dt}\|e\|^2=e^T(A+A^T)e
=-q^2e_x^2-(q^2+1)e_z^2
\leq-q^2\|e\|^2.
$$
For $q\ne0$ the derivative is strictly negative whenever $e\ne0$. Integrating the inequality proves the <symmetric-part contraction criterion> here explicitly:
$$
\boxed{\|s(t)-s_*\|\leq e^{-q^2t/2}\|s(0)-s_*\|.}
$$
Thus every nonstationary initial state approaches the unique <steady state>, with strictly decreasing distance; the <steady state> itself has constant zero distance. If the decoupled $y$ coordinate is retained, it contributes $-e_y^2$ to the squared-distance derivative and also decays, so the conclusion holds for the full qubit state.
For the target-dependent settings in (e), $q=-\cos\theta_d$. Every nonequatorial target is therefore globally attractive. At an equatorial target, $q=\alpha=0$: $x(t)=x(0)$, $z(t)=z(0)e^{-t/2}$, and $y(t)=y(0)e^{-t/2}$. A general state approaches $(x(0),0,0)$ rather than the specified pure target $x=\pm1$. This proves the equatorial exception, not merely failure of a sufficient estimate.
The parameter dependence matters: $q=0$ alone does not exclude attractivity for every possible driving strength. If $\alpha\ne0$, the reduced matrix has negative <trace> and positive <determinant> $4\alpha^2$, so both <eigenvalues> have negative real parts even though its symmetric part is only semidefinite. The nonattractive case asserted for the equatorial target also uses its prescribed $\alpha=0$.
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