= Solution
Choose a normal ball on which $\exp_p$ is a diffeomorphism, and take $\varepsilon$ smaller than half its radius. For $|X|<\varepsilon$, $t\mapsto\exp_p(tX)$ is a geodesic of length $|X|$. If a competing curve remains in the normal ball, write it in polar form. The <Gauss lemma> makes radial and angular velocities orthogonal, so its speed is at least the absolute radial speed and its length is at least $|X|$. A curve leaving the larger normal ball already accumulates more than $|X|$ in radial variation. Thus the radial geodesic minimizes length.
At $q$, both $X$ and the geodesic sphere $\Sigma$ are hypersurfaces. If their tangent hyperplanes were distinct, they would be transverse. The <transverse intersection theorem> would then make $X\cap\Sigma$ a submanifold of dimension
$$
(m-1)+(m-1)-m=m-2.
$$
For $m\geq3$ this has positive dimension near $q$, contradicting that the intersection is the singleton $\{q\}$. Therefore $T_qX=T_q\Sigma$.
The conclusion fails in dimension two because a transverse intersection is zero-dimensional and may be isolated. In the Euclidean plane, let $\Sigma$ be the unit circle, $q=(1,0)$, and let $X=\{(x,0):1/2<x<3/2\}$. Then $X\cap\Sigma=\{q\}$, but $T_qX$ is horizontal whereas $T_q\Sigma$ is vertical.
Solved by gpt-5.6-sol high.
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