= Solution
First solve the hinted fixed-gap problem. Let $y$ measure distance across the gap and $H(x)=h+x^2/(2a)$ be its leading local height. The <lubrication approximation> gives $\mu u_{yy}=p_x$, with $u=0$ at $y=0,H$, so
$$
u=-\frac{p_x}{2\mu}y(H-y),\qquad q=\int_0^H u\,dy=-\frac{H^3p_x}{12\mu}.
$$
The constant flux per unit transverse length is $q$. Integrating its <pressure> <gradient> gives the <pressure-driven flux through a parabolic lubrication gap>:
$$
\Delta p=12\mu q\int_{-\infty}^{\infty}\frac{dx}{(h+x^2/(2a))^3}.
$$
With $x=\sqrt{2ah}\,t$, the <integral> is $\sqrt{2a}\,h^{-5/2}\int_{-\infty}^{\infty}(1+t^2)^{-3}dt$. The substitution $t=\tan\chi$ gives $\int_{-\pi/2}^{\pi/2}\cos^4\chi\,d\chi=3\pi/8$, and therefore
$$
\boxed{\Delta p=\frac{9\pi\mu q\sqrt{2a}}{2h^{5/2}}.}
$$
Now take the <sphere> speed to be $U>0$ downwards and assume zero net liquid through-flow, so the displaced liquid must pass through the annular gap. Near the equator, the <sphere>'s surface radius is $R-z^2/(2R)+\cdots$, giving $H(z)=d+z^2/(2R)$. In the <sphere> frame the relative <volume flux> is $U\pi(R+d)^2$; dividing by circumference $2\pi R$ gives $q\simeq UR/2$. The Couette contribution from the moving tube wall is only order $Ud$, smaller by $d/R$, so the fixed-gap <pressure> result gives
$$
\Delta p\simeq\frac{9\pi\mu UR\sqrt{2R}}{4d^{5/2}}.
$$
The <pressure> changes principally within axial distance $\sqrt{Rd}$ around the equator. It is nearly constant on each hemisphere, so its leading <force> is the <pressure> difference times projected area, $F_p\simeq\pi R^2\Delta p$. Direct viscous shear contributes a relative order $d/R$ and is smaller. Balance with the buoyant weight:
$$
\frac{4\pi R^3}{3}\Delta\rho g\simeq\pi R^2\frac{9\pi\mu UR\sqrt{2R}}{4d^{5/2}}.
$$
Thus the <settling speed of a nearly occluding sphere without through-flow> is
$$
\boxed{U\sim\frac{8\sqrt2\,\Delta\rho g d^2}{27\pi\mu}\left(\frac dR\right)^{1/2}.}
$$
The printed speed lacks the factor $1/\pi$. The parabolic-gap <integral> above fixes the normalization: its $3\pi/8$ cannot be dropped. With standard <dynamic viscosity> and <Stokes drag law> conventions, \b[the requested coefficient must be corrected by dividing the printed expression by $\pi$]. Also, the bypass result assumes no net through-flow; allowing a piston-like displacement of the entire fluid column would require specifying tube end conditions and a different resistance balance.
A <sphere> of radius $d$ in an unbounded fluid has sedimentation speed $U_d=2\Delta\rho gd^2/(9\mu)$. Therefore
$$
\boxed{\frac U{U_d}\sim\frac{4\sqrt2}{3\pi}\left(\frac dR\right)^{1/2}\ll1.}
$$
The large confined <sphere> falls much more slowly even than this much smaller unconfined <sphere>.
For radial offset $e<d$, let $\epsilon=e/d$. To leading order the narrowest gap varies as $d(\phi)=d(1+\epsilon\cos\phi)$. All circumferential strips share the same <pressure> drop and act as parallel channels. Inverting the gap-resistance formula shows that each carries flux proportional to $d(\phi)^{5/2}\Delta p$. At fixed buoyant load, the leading <pressure> drop is fixed, so the speed ratio is
$$
\boxed{\frac{U(e)}{U(0)}\simeq\frac1{2\pi}\int_0^{2\pi}(1+\epsilon\cos\phi)^{5/2}d\phi>1\quad(0<\epsilon<1).}
$$
The <eccentricity increases narrow-gap bypass mobility> result holds because the gain through the wider side outweighs the loss through the narrower side. Strict <convexity> of $x^{5/2}$ and the zero mean of cosine prove the inequality. Pairing the positive- and negative-cosine contributions in the derivative proves monotonicity with offset. The broader hint's assertion for every exponent $\alpha>0$ is not correct for $0<\alpha<1$, when <concavity> reverses the inequality; the required exponent $5/2$ is in the valid convex range.
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