Shot-noise variance of gradient estimators

The one-qubit circuit $R_Y(\theta)\lvert0\rangle$ with $R_Y(\theta)=e^{-i\theta Y/2}$ is measured in the computational basis, and each shot returns $z=+1$ (outcome $0$) or $z=-1$ (outcome $1$). The model output is

$f(\theta)=\langle Z\rangle=\cos\theta .$

An evaluation $\hat f(\varphi)$ runs the circuit at angle $\varphi$ for $N=1000$ fresh shots and returns the mean of the $N$ outcomes; different evaluations use independent shots. Consider the two gradient estimators

$\hat g_{\mathrm{FD}}=\frac{\hat f(\theta+h)-\hat f(\theta-h)}{2h},\qquad\hat g_{\mathrm{PSR}}=\frac12\Big[\hat f\big(\theta+\tfrac\pi2\big)-\hat f\big(\theta-\tfrac\pi2\big)\Big].$

At $\theta=\pi/4$ and $N=1000$, compute

(a) the variance $\mathrm{Var}[\hat g_{\mathrm{FD}}]$ of the central finite-difference estimator with $h=0.01$;

(b) the variance $\mathrm{Var}[\hat g_{\mathrm{PSR}}]$ of the parameter-shift estimator.

Enter two real numbers in this order. Each value is graded with relative tolerance $10^{-3}$ (plus an absolute $10^{-7}$). You may type decimals such as 2.5e-4 or exact expressions such as 1/(40.01^21000); an expression must be correct to within $10^{-6}$.