I figured out that the teacher wanted us to use the multivariate delta method:

Part (b)

We have $$T=\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n 1\{X_i>0\}}=\dfrac{n\bar{X}_n}{n-nZ_n} = \dfrac{\bar{X}_n}{1-Z_n}$$Let $h(x,y)=x/(1-y)$ with gradient evaluated at $(\theta, e^{-\theta})'$ $ \mathbf{D} = \nabla h(\theta, e^{-\theta}) = \left(\dfrac{1}{1-e^{-\theta}}, \dfrac{\theta}{(1-e^{-\theta})^2}\right)' $. Then by the multivariate delta method $$\sqrt{n}\left( \dfrac{\bar{X}}{1-Z_n} - \dfrac{\theta}{1-e^{-\theta}}\right) \overset{D}{\longrightarrow} W, \quad W\sim \text{N}\left( 0, \mathbf{D}\Sigma \mathbf{D}'\right)$$ where $$ \mathbf{D}\Sigma \mathbf{D}' = \begin{pmatrix}\dfrac{1}{1-e^{-\theta}} & \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix}\begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}(1-e^{-\theta})\end{pmatrix}\begin{pmatrix}\dfrac{1}{1-e^{-\theta}} \\ \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix} = \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3}$$

Part (c)

$\begin{align*} \text{E}[X|X\geq 1] &= \dfrac{\sum_{k=1}^\infty kP(X=k)}{P(X\geq 1)}\\ &= \dfrac{\sum_{k=0}^\infty kP(X=k) - 0*P(X=0)}{1-P(X=0)}\\ &= \dfrac{\theta}{1-e^{-\theta}}\\ \text{Var}[X|X\geq 1] &= \text{E}[X^2|X\geq 1] - (\text{E}[X|X\geq 1])^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \left(\dfrac{\theta}{1-e^{-\theta}}\right)^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \dfrac{\theta^2}{(1-e^{-\theta})^2}\\ &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ \dfrac{\text{aVar}[T]}{\text{Var}[X|X\geq 1]} &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3} \Bigg/ \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ &= \dfrac{1}{1-e^{-\theta}} < 1 \end{align*}$ The asymptotic variance of the estimator in part (b) is smaller than the exact variance. So the asymptotic estimator is less efficient.

I figured out that the teacher wanted us to use the multivariate delta method:

Part (b)

We have $$T=\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n 1\{X_i>0\}}=\dfrac{n\bar{X}_n}{n-nZ_n} = \dfrac{\bar{X}_n}{1-Z_n}$$Let $h(x,y)=x/(1-y)$ with gradient evaluated at $(\theta, e^{-\theta})'$ $ \mathbf{D} = \nabla h(\theta, e^{-\theta}) = \left(\dfrac{1}{1-e^{-\theta}}, \dfrac{\theta}{(1-e^{-\theta})^2}\right)' $. Then by the multivariate delta method $$\sqrt{n}\left( \dfrac{\bar{X}}{1-Z_n} - \dfrac{\theta}{1-e^{-\theta}}\right) \overset{D}{\longrightarrow} W, \quad W\sim \text{N}\left( 0, \mathbf{D}\Sigma \mathbf{D}'\right)$$ where $$ \mathbf{D}\Sigma \mathbf{D}' = \begin{pmatrix}\dfrac{1}{1-e^{-\theta}} & \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix}\begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}(1-e^{-\theta})\end{pmatrix}\begin{pmatrix}\dfrac{1}{1-e^{-\theta}} \\ \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix} = \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3}$$

Part (c)

$\begin{align*} \text{E}[X|X\geq 1] &= \dfrac{\sum_{k=1}^\infty kP(X=k)}{P(X\geq 1)}\\ &= \dfrac{\sum_{k=0}^\infty kP(X=k) - 0*P(X=0)}{1-P(X=0)}\\ &= \dfrac{\theta}{1-e^{-\theta}}\\ \text{Var}[X|X\geq 1] &= \text{E}[X^2|X\geq 1] - (\text{E}[X|X\geq 1])^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \left(\dfrac{\theta}{1-e^{-\theta}}\right)^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \dfrac{\theta^2}{(1-e^{-\theta})^2}\\ &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ \dfrac{\text{aVar}[T]}{\text{Var}[X|X\geq 1]} &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3} \Bigg/ \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ &= \dfrac{1}{1-e^{-\theta}} < 1 \end{align*}$ The asymptotic variance of the estimator in part (b) is smaller than the exact variance