= Solution
Using the <Levi-Civita symbol> and the <epsilon-delta identity>,
$$
[\nabla\times(\nabla\times A)]_i
=\epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lA_m
=\partial_i(\partial_jA_j)-\partial_j\partial_jA_i.
$$
Therefore
$$
\boxed{\nabla\times(\nabla\times A)=\nabla(\nabla\mathbin\cdot A)-\nabla^2A}.
$$
One particularly simple <vector potential> for $u$ is
$$
\boxed{A=\frac13(z^3,x^3,y^3)}.
$$
Direct differentiation gives $\nabla\mathbin\cdot A=0$ and $\nabla\times A=(y^2,z^2,x^2)=u$.
Parametrize the curved cone by
$$
R(z,\varphi)=(z\tan\alpha\cos\varphi,
z\tan\alpha\sin\varphi,z).
$$
The outward <oriented surface element> is
$$
d\mathbf S=(R_\varphi\times R_z)\,d\varphi\,dz
=z(\tan\alpha\cos\varphi,\tan\alpha\sin\varphi,-\tan^2\alpha)\,d\varphi\,dz.
$$
Substitution and integration over $0\leq z\leq h$, $0\leq\varphi<2\pi$ give
$$
\boxed{\int_{S_1}u\mathbin\cdot d\mathbf S
=-\frac{\pi h^4\tan^4\alpha}{4}}.
$$
Because $\nabla\mathbin\cdot u=0$, the <divergence theorem> predicts that the two outward fluxes sum to zero. On the top disk $S_2$, $d\mathbf S=(0,0,1)\,dx\,dy$ and $u_z=x^2$, so, with $R=h\tan\alpha$,
$$
\int_{S_2}u\mathbin\cdot d\mathbf S
=\int_{x^2+y^2\leq R^2}x^2\,dx\,dy
=\boxed{\frac{\pi R^4}{4}}
=\frac{\pi h^4\tan^4\alpha}{4},
$$
as predicted.
The outward orientation on $S_2$ induces the stated anticlockwise orientation on $C$, whereas the outward orientation on $S_1$ induces the reverse orientation. Since $u=\nabla\times A$, <Stokes theorem> predicts
$$
\int_{S_2}u\mathbin\cdot d\mathbf S=\oint_CA\mathbin\cdot d\mathbf l,
\qquad
\int_{S_1}u\mathbin\cdot d\mathbf S=-\oint_CA\mathbin\cdot d\mathbf l.
$$
On $C$, put $(x,y,z)=(R\cos\varphi,R\sin\varphi,h)$. Then
$$
\oint_CA\mathbin\cdot d\mathbf l
=\frac13\int_0^{2\pi}
\left[-h^3R\sin\varphi+R^4\cos^4\varphi\right]d\varphi
=\boxed{\frac{\pi R^4}{4}},
$$
verifying both identities.
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