$ f(x|\theta)=2x\theta e^{-\theta x^2}I_{(0,\infty)}(x)\hspace{0.5cm}\theta\in\mathbb{R}^+ $UMVUE for $h(\theta)=1/\theta$ and $h(\theta)=\theta$

Question

Let $X$ a random sample with density function $$f(x|\theta)=2x\theta e^{-\theta x^2}I_{(0,\infty)}(x)\hspace{0.5cm}\theta\in\mathbb{R}^+$$

Then $$f_\theta(\vec{x})=\left(2^n\prod _{i=1}^n x_i\right)\theta^n\exp{-\theta\sum_{i=1}^n x_i^2}\hspace{0.5cm}\underset{\forall i=1,\dots,n}{x_i,\theta\in\mathbb{R}^+}$$ It belongs to the exponential family and $$T(X)=\sum\limits_{i=1}^n X_i^2$$ is a minimal statistic for $\theta$

Question 1): $$\frac{d}{d\theta}\ln f_\theta(\vec{x})=-\left[T-\frac{n}{\theta}\right]=-n\left[\frac{T}{n}-\frac{1}{\theta}\right]$$ Then $T$ is an efficent statistic for $h(\theta)=\frac{1}{\theta}$, so is T the UMVUE for $1/\theta?$

Question 2) How can I find the UMVUE for $\theta $ ?

Thank you