= Solution
For a scalar $\phi$, the displayed basis identities give
$$
\nabla\phi=\mathbf e_r\phi_r+\mathbf e_\theta\frac1r\phi_\theta.
$$
Taking the <divergence> and differentiating the basis vectors yields the <Laplacian in polar coordinates>
$$
\boxed{\nabla^2\phi=\frac1r\frac{\partial}{\partial r}\left(r\frac{\partial\phi}{\partial r}\right)+\frac1{r^2}\frac{\partial^2\phi}{\partial\theta^2}}.
$$
Consequently,
$$
\nabla^2\!\left(\alpha r^\beta\cos(\gamma\theta)\right)
=\boxed{\alpha(\beta^2-\gamma^2)r^{\beta-2}\cos(\gamma\theta)}.
$$
The regular harmonic mode with angular dependence $\cos2\theta$ is $Cr^2\cos2\theta$. Its radial derivative at $r=a$ is $2Ca\cos2\theta$, so $C=1/(2a)$. A constant is invisible to the <Neumann boundary condition>, giving
$$
\boxed{\phi(r,\theta)=\frac{r^2}{2a}\cos2\theta+C_0.}
$$
If two solutions existed, their difference $u$ would satisfy $\nabla^2u=0$ and $\partial_nu=0$. <Green's first identity> gives $\int_D|\nabla u|^2=\int_{\partial D}u\,\partial_nu-\int_Du\nabla^2u=0$, so $\nabla u=0$ and $u$ is constant. Thus these are all the solutions.
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