= Solution
Under the <first-order smoothing approximation>, neglect the fluctuating product in the <resistive induction equation> while retaining its contribution to the mean electromotive force. Since the test field $\mathbf B_0$ is constant and the velocity is a <solenoidal vector field>, the fluctuation equation is
$$
\partial_t\mathbf b-\eta\nabla^2\mathbf b=(\mathbf B_0\cdot\nabla)\mathbf u.
$$
Let $a=\eta k^2>0$, $d=a^2+\Omega^2$, and $\mathbf C_j=i(\mathbf B_0\cdot\mathbf k_j)\hat{\mathbf u}_j$. For each <Fourier mode>, the <diffusion equation> becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega\ne0$. For the first mode,
$$
a\mathbf p_1+\Omega\mathbf q_1=\mathbf C_1,\qquad a\mathbf q_1-\Omega\mathbf p_1=0,
$$
while for the second,
$$
a\mathbf p_2+\Omega\mathbf q_2=0,\qquad a\mathbf q_2-\Omega\mathbf p_2=\mathbf C_2.
$$
Thus \b[the complex response coefficients] are
$$
\boxed{\mathbf p_1=\frac{a\mathbf C_1}{d},\quad \mathbf q_1=\frac{\Omega\mathbf C_1}{d},\quad
\mathbf p_2=-\frac{\Omega\mathbf C_2}{d},\quad \mathbf q_2=\frac{a\mathbf C_2}{d}.}
$$
They give
$$
\mathbf b=\operatorname{Re}\sum_{j=1}^{2}e^{i\mathbf k_j\cdot\mathbf x}(\mathbf p_j\cos\Omega t+\mathbf q_j\sin\Omega t).
$$
An arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\operatorname{Re}\sum_j\mathbf r_j e^{i\mathbf k_j\cdot\mathbf x-a t}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final <real part>.
To justify the helicity relation, write $\hat{\mathbf u}_j=\mathbf a_j+i\mathbf c_j$ with real vectors. Solenoidality gives $\mathbf k_j\cdot\mathbf a_j=\mathbf k_j\cdot\mathbf c_j=0$. Their <cross product> is parallel to $\mathbf k_j$, and
$$
\frac12\operatorname{Re}(i\hat{\mathbf u}_j^*\times\hat{\mathbf u}_j)
=-\mathbf a_j\times\mathbf c_j=H_j\mathbf k_j,
\qquad
H_j=\frac{\mathbf k_j\cdot\operatorname{Re}(i\hat{\mathbf u}_j^*\times\hat{\mathbf u}_j)}{2k^2}\in\mathbb R.
$$
This is the <helicity vector of a solenoidal Fourier mode>. In these conventions the spatially averaged <kinetic helicity density> of $\operatorname{Re}(\hat{\mathbf u}_je^{i\mathbf k_j\cdot\mathbf x})$ is $-H_jk^2$; the sign is set by the specified cross-product order.
For equal <wavevectors>, spatial averaging of two real harmonic fields gives
$$
\left\langle\operatorname{Re}(\mathbf v e^{i\mathbf k\cdot\mathbf x})\times\operatorname{Re}(\mathbf w e^{i\mathbf k\cdot\mathbf x})\right\rangle_x
=\frac12\operatorname{Re}(\mathbf v^*\times\mathbf w).
$$
Time averaging gives another factor $1/2$, because $\langle\cos^2\Omega t\rangle_t=\langle\sin^2\Omega t\rangle_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the <mean-field electromotive force>
$$
\mathcal E_j=\frac{a}{4d}(\mathbf B_0\cdot\mathbf k_j)\operatorname{Re}(i\hat{\mathbf u}_j^*\times\hat{\mathbf u}_j)
=\frac{a}{2d}H_j\mathbf k_j(\mathbf k_j\cdot\mathbf B_0).
$$
Distinct modes with $\mathbf k_1\ne\pm\mathbf k_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are $\mathbf p_2=-\Omega\mathbf C_2/d$ and $\mathbf q_1=\Omega\mathbf C_1/d$, while integration by parts gives $\langle\mathbf u_1\times(\mathbf B_0\cdot\nabla)\mathbf u_2\rangle_x=\langle\mathbf u_2\times(\mathbf B_0\cdot\nabla)\mathbf u_1\rangle_x$ for the real spatial fields. This handles all the distinct <wavevectors> allowed by the question, assuming the usual periodic-cell or whole-space spatial average.
Thus \b[the <alpha tensor> is]
$$
\boxed{\alpha_{ij}=\frac{\eta k^2}{2(\Omega^2+\eta^2k^4)}\left(H_1 k_{1i}k_{1j}+H_2 k_{2i}k_{2j}\right),\qquad \boldsymbol{\mathcal E}=\boldsymbol\alpha\mathbf B_0.}
$$
Both averaging factors matter. The <alpha tensor> is real and symmetric, and this contribution vanishes when the mode <kinetic helicity densities> vanish.
For the two perpendicular wavevectors and common $H_1=H_2=H$, put $\alpha_0=\eta H k^4/[2(\Omega^2+\eta^2k^4)]$. Then
$$
\boxed{\boldsymbol\alpha=\alpha_0\operatorname{diag}(1,1,0).}
$$
\b[<Mean-field dynamo> action is possible when $\alpha_0\ne0$ and sufficiently long mean-field wavelengths are allowed.] The missing $zz$ entry does not prevent an <alpha-squared dynamo>. Indeed a slowly varying mean <magnetic field> transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeys
$$
\partial_t\overline{\mathbf B}=\nabla\times(\boldsymbol\alpha\overline{\mathbf B})+\eta\nabla^2\overline{\mathbf B},
\qquad
\lambda\begin{pmatrix}\overline B_x\\\overline B_y\end{pmatrix}
=\begin{pmatrix}-\eta K^2&-i\alpha_0K\\i\alpha_0K&-\eta K^2\end{pmatrix}\begin{pmatrix}\overline B_x\\\overline B_y\end{pmatrix}.
$$
The two growth rates are $-\eta K^2\pm\alpha_0K$, so the larger is positive if $0<|K|<|\alpha_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero <curl> of its <mean-field electromotive force> and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular <alpha effect> does not yield growth. The conclusion concerns the stated <first-order smoothing approximation>; no uncomputed turbulent-diffusion correction is assumed.
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