= Solution
Let $\mathbf n=\mathbf X/R$, $\zeta=6\pi\mu a$, and let
$$
\mathsf G(\mathbf r)=\frac{I+\widehat{\mathbf r}\widehat{\mathbf r}}{8\pi\mu r},\qquad\mathsf A=\zeta\mathsf G(\mathbf X)=\frac{3a}{4R}(I+\mathbf n\mathbf n).
$$
Use the <method of reflections for Stokes flow>. At the fixed <sphere>, the first <sphere> produces the incident <Stokeslet> <velocity> $\mathbf u_\infty(0)=\mathsf A\mathbf V$ to leading order. In <Faxén translation law>, set the second <sphere>'s translational <velocity> to zero. The applied holding <force> is therefore
$$
\boxed{\mathbf F_2=-\frac{9\pi\mu a^2}{2R}\left[\mathbf V+(\mathbf V\cdot\mathbf n)\mathbf n\right]+O\left(\frac{\mu Va^4}{R^3}\right).}
$$
The first <sphere>'s finite-radius <potential dipole> adds $O(Va^3/R^3)$ to the incident <velocity>, as does the <Laplacian> term in <Faxén translation law> at the fixed <sphere>. Multiplying by $\zeta$ gives the stated next correction. There is no intermediate $O(\mu Va^3/R^2)$ term. This is the <holding force and torque for a sphere in a distant Stokeslet>.
The <vorticity> of a <Stokeslet> at displacement $\mathbf r$ is $\mathbf F\times\mathbf r/(4\pi\mu r^3)$. At the fixed <sphere>, $\mathbf r=-\mathbf X$, so
$$
\boldsymbol\omega_\infty(0)=-\frac{3a}{2R^3}\mathbf V\times\mathbf X.
$$
Set its <angular velocity> to zero in <Faxén rotation law>. The applied holding couple is
$$
\boxed{\mathbf G_2=-4\pi\mu a^3\boldsymbol\omega_\infty(0)=\frac{6\pi\mu a^4}{R^3}\mathbf V\times\mathbf X+O\left(\frac{\mu Va^6}{R^4}\right).}
$$
The displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the <sphere>.
The leading reflected flow at the first <sphere> is $\mathsf G(\mathbf X)\mathbf F_2=-\mathsf A^2\mathbf V$. Apply <Faxén translation law> there with its prescribed <force> $\mathbf F=\zeta\mathbf V$. Since $(I+\mathbf n\mathbf n)^2=I+3\mathbf n\mathbf n$, the <mobility correction from a fixed distant sphere> is
$$
\boxed{\dot{\mathbf X}=\mathbf V-\frac{9a^2}{16R^2}\left[\mathbf V+3(\mathbf V\cdot\mathbf n)\mathbf n\right]+O\left(\frac{Va^4}{R^4}\right).}
$$
The next correction comes from finite-radius terms in the incident/reflected flow and in <Faxén translation law>, together with the fixed <sphere>'s induced <stresslet> and holding-couple <rotlet>. Each gives $O(Va^4/R^4)$ at the first <sphere>. The symbol $\mathbf V$ is its isolated-sphere <velocity> scale, not its actual <velocity> in the two-sphere problem.
The first <sphere> has no applied couple. Its <angular velocity> follows from <Faxén rotation law> and the reflected <Stokeslet>:
$$
\boldsymbol\Omega_1=\frac{\mathbf F_2\times\mathbf X}{8\pi\mu R^3}+O\left(\frac{Va^4}{R^5}\right)=\boxed{-\frac{9a^2}{16}\frac{\mathbf V\times\mathbf X}{R^4}+O\left(\frac{Va^4}{R^5}\right).}
$$
The component of $\mathbf F_2$ parallel to $\mathbf X$ contributes no <vorticity>. The next reflected <rotlet> and <stresslet> give the stated error order.
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