= Solution
Use the unwrapped angle on the real line. Integrating its <Langevin equation> gives $\theta(t)-\theta(0)=\omega t+\eta(t)$, where $\eta=\int_0^t\beta(s)ds$ is a centered <Gaussian random variable> with variance $2D_rt$. Therefore
$$
\boxed{\langle[\theta(t)-\theta(0)]^2\rangle=\omega^2t^2+2D_rt.}
$$
A wrapped angular difference is bounded and would not satisfy this formula.
For the orientation correlation, $\mathbf n(t)\cdot\mathbf n(0)=\cos[\theta(t)-\theta(0)]$. The <characteristic function> gives $\langle e^{i\eta}\rangle=e^{-\langle\eta^2\rangle/2}=e^{-D_rt}$, so
$$
\langle\mathbf n(t)\cdot\mathbf n(0)\rangle=e^{-D_rt}\cos\omega t.
$$
The translational noise is independent of the orientation, hence the cross terms vanish. Summing its two component correlations gives $\langle\mathbf u(t)\cdot\mathbf u(0)\rangle=4D\delta(t)$. Consequently, for $t\ge0$,
$$
\boxed{\langle\mathbf v(t)\cdot\mathbf v(0)\rangle=4D\delta(t)+v^2e^{-D_rt}\cos\omega t.}
$$
The extension to negative time uses $e^{-D_r|t|}$. Because the position contains <Brownian motion>, the instantaneous noise velocity and its correlation are <distributions>; the delta contribution is not a finite ordinary velocity at $t=0$.
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