= Solution
{c}
Use the <Hellinger distance> convention in the original PDF:
$$
d_{\mathrm{Hell}}(\mu,\mu')^2=\frac12\int(\sqrt p-\sqrt q)^2\,d\nu
=1-\int\sqrt{pq}\,d\nu,
\qquad p=\frac{d\mu}{d\nu},\quad q=\frac{d\mu'}{d\nu}.
$$
The TeX transcription incorrectly places the integral outside the outer square root in the definition; the displayed formula for the squared distance and the original PDF both use the convention above. Although <Lebesgue measure> is not a probability measure, we may use it as the common dominating measure: the integral is unchanged by replacing it with any equivalent dominating <probability measure>.
For unit-variance <normal distributions>, put $\bar\theta=(\theta_1+\theta_2)/2$ and $\Delta=\theta_1-\theta_2$. The product of the square roots of their <probability density functions> is
$$
\sqrt{p_1(x)p_2(x)}
=\frac1{\sqrt{2\pi}}\exp\left[-\frac{(x-\theta_1)^2+(x-\theta_2)^2}{4}\right]
=e^{-\Delta^2/8}\frac1{\sqrt{2\pi}}e^{-(x-\bar\theta)^2/2}.
$$
The final factor integrates to one, so
$$
\boxed{d_{\mathrm{Hell}}(\mu_1,\mu_2)^2=1-e^{-(\theta_1-\theta_2)^2/8}.}
$$
For the bound at arbitrary $\sigma_1,\sigma_2>0$, use the given formula for the <Hellinger distance between normal distributions> and write
$$
S=\sigma_1^2+\sigma_2^2,\qquad
B=\sqrt{\frac{2\sigma_1\sigma_2}{S}}\leq1,\qquad
x=\frac{(\theta_1-\theta_2)^2}{4S}.
$$
Since $1-e^{-x}\leq x$ for $x\geq0$,
$$
\begin{aligned}
d_{\mathrm{Hell}}^2
&=(1-B)+B(1-e^{-x})\\
&\leq\frac{1-B^2}{1+B}+x\\
&=\frac{(\sigma_1-\sigma_2)^2}{S(1+B)}
+\frac{(\theta_1-\theta_2)^2}{4S}\\
&\leq\frac{(\sigma_1-\sigma_2)^2+(\theta_1-\theta_2)^2}{S}.
\end{aligned}
$$
Therefore one valid choice is $\boxed{C_{\sigma_1,\sigma_2}=1/(\sigma_1^2+\sigma_2^2)}$. In particular this bound is uniform when the two standard deviations stay bounded away from zero.
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