= Solution
There is a genuine normalization error in the printed function. Write $Y_i\sim\operatorname{Unif}(0,2)$ independently, $g(y)=1/(\pi(1+y^2))$ and $H(y)=y^2-4/3$. The internal sum produces a scalar; taking its mean does not divide by $n$. The literal returned value is therefore
$$
T_{\beta,\mathrm{printed}}=\frac12-2\sum_{i=1}^n\{g(Y_i)-\beta H(Y_i)\}.
$$
Since $\mathbb EH=0$ and $\mathbb Eg=\arctan(2)/(2\pi)=1/4-\theta/2$, its expectation is
$$
\mathbb E T_{\beta,\mathrm{printed}}=n\theta+\frac{1-n}{2}.
$$
It does not estimate $\theta$ unbiasedly unless $n=1$, and no fixed <control variate> coefficient removes this bias. In particular the actual call with $\beta=0$ returns
$$
\boxed{\check\theta_{\mathrm{printed}}=\frac12-2\sum_{i=1}^n\frac1{\pi(1+Y_i^2)}.}
$$
This is an instance of <summing before taking a mean changes Monte Carlo normalization>.
The intended corrected method replaces the sum by a <sample mean>:
$$
T_\beta=\frac12-\frac2n\sum_{i=1}^n\{g(Y_i)-\beta H(Y_i)\}.
$$
It uses symmetry to write the tail as $\theta=1/2-\int_0^2g(y)\,dy$, then <Monte Carlo integration> on a bounded interval, and finally a mean-zero <control variate>. For $\beta=0$ the intended estimator is $1/2-(2/n)\sum_i g(Y_i)$, whose summands are smooth and bounded. Keeping the average normalization, $\mathbb ET_\beta=\theta$ and
$$
\operatorname{Var}(T_\beta)=\frac4n\{\operatorname{Var}(g)-2\beta\operatorname{Cov}(g,H)+\beta^2\operatorname{Var}(H)\}.
$$
Differentiating this quadratic gives the <Cauchy tail control variate on a bounded interval> coefficient
$$
\beta_* =\frac{\operatorname{Cov}(g,H)}{\operatorname{Var}(H)}.
$$
Here $\mathbb EY^2=4/3$, $\mathbb EY^4=16/5$, so $\operatorname{Var}(H)=64/45$. Also
$$
\operatorname{Cov}(g,H)=\frac1{2\pi}\int_0^2\frac{y^2-4/3}{1+y^2}\,dy=\frac1\pi-\frac{7\arctan2}{6\pi}=\frac1\pi-\frac7{12}+\frac{7\theta}{6}.
$$
Consequently
$$
\boxed{\beta_* =\frac{45}{64}\left(\frac1\pi-\frac7{12}+\frac{7\theta}{6}\right)\approx-0.06528.}
$$
The negative sign is consistent with $g$ decreasing and $H$ increasing. One may estimate the <covariance> and <variance> using a separate pilot sample and fix the resulting coefficient for a fresh sample. This coefficient minimizes the literal function's <variance> as well, but its missing factor of $1/n$ still leaves the bias above.
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