= Solution
The almost-sure bound gives $\tilde\theta_n-\hat\theta_n=o_{\mathbb P}(n^{-1/2})$, so <Slutsky theorem> transfers both consistency and the asymptotic normal law to $\tilde\theta_n$. The <continuous mapping theorem> and <delta method> then give
$$
\boxed{\psi(\tilde\theta_n)\xrightarrow{\mathbb P}\psi(\theta_0),\qquad
\sqrt n\bigl(\psi(\tilde\theta_n)-\psi(\theta_0)\bigr)
\xrightarrow{d}\mathcal N\!\left(0,\frac{\psi'(\theta_0)^2}{I(\theta_0)}\right).}
$$
If $\psi'(\theta_0)=0$, the latter is the <Dirac measure> at zero, conventionally a degenerate <normal distribution>; a nondegenerate second-order <limit> would need a different normalization and extra smoothness.
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