We have the joint pdf
$$
f(\vec x ; \theta) = \theta^n c^{\theta n} \prod_{i=1}^n x_i^{-(\theta+1)}\mathbb{1}_{x_i \ge c}
=\mathbb{1}_{x_{(1)} \ge c} \left[ \theta^n c^{\theta n} \right] \exp \left[
-(\theta+1) \sum_{i=1}^n \ln x_i\right]
$$
and so by the Exponential-Family factorization $\sum_{i=1}^n \ln X_i$ is complete & sufficient for the distribution.
For a preliminary result, consider $Y = \ln(X) - \ln(c)$ where $X$ follows from the given pareto distribution i.e. $f_X(x) = \theta c^\theta x^{-(\theta+1)} \mathbb{1}_{x \ge c}$. Then, since $X = ce^Y$, we get
$$
f_Y(y) = f_X(ce^Y) \left\vert \frac{dx}{dy} \right\vert
= \theta c^\theta (c e^Y)^{-(\theta+1)} \mathbb{1}_{ ce^Y \ge c} \cdot ce^Y = \theta e^{-y \theta} \mathbb{1}_{ y \ge 0}
$$
which is the pdf of an exponential rate = $\theta$ distribution. Define $Y_i := \ln(X_i) - \ln(c)$
It follows that $\sum_{i=1}^n Y_i = \sum_{i=1}^n (\ln X_i - \ln (c))$ follows a $\Gamma(n,\theta)$ distribution since it's the sum of $n$ independent exponential rate $\theta$ random variables. Note that the mean of an exponential rate $\theta$ r.v. is $1/\theta$ and the mean of a $\Gamma(n,\theta)$ r.v. is $n/\theta$.
So $\frac{1}{n} \sum_{i=1}^n Y_i$ is an unbiased estimator of $1/\theta$, and it's natural to guess that $1/ \left( \frac{1}{n} \sum_{i=1}^n Y_i \right)$ is an unbiased estimator of $\theta$.
Let $Z \sim \Gamma(n,\theta)$. Then, the expecation of $1/ \left( \frac{1}{n} \sum_{i=1}^n Y_i \right)$ equals:
\begin{align*}
E \left[ \frac{n}{Z} \right] &= n \int_0^\infty \frac{1}{z} \frac{1}{\Gamma(n)} \theta^n z^{n-1} e^{- \theta z} \; dz \\
&= n \int_0^\infty \frac{1}{\Gamma(n)} \theta^n z^{n-2} e^{- \theta z} \; dz \\
&= n \frac{ \theta \Gamma(n-1)}{\Gamma(n)} \int_0^\infty \frac{1}{\Gamma(n-1)} \theta^{n-1} z^{n-2} e^{- \theta z} \; dz
\end{align*}
and this equals $\theta n \dfrac{ (n-2)!}{(n-1)!}= \frac{n}{n-1} \theta$ since the rightmost integral is integrating the pdf of a $\Gamma(n-1,\theta)$ random variable over its support.
It follows from Lehmann Scheffe that $\dfrac{n-1}{n} \cdot \dfrac{1}{\frac{1}{n} \sum_{i=1}^n Y_i} = \dfrac{n-1}{\sum_{i=1}^n (\ln X_i - \ln c) }$ is the UMVUE of $\theta$.
To answer your first question, even if an unbiased estimator exists it does not guarantee that a UMVUE exists.
Consider a single observation $X$ having the uniform distribution on $(\theta,\theta+1)$ and suppose we have to estimate $g(\theta)$ for some function $g$.
So $X$ is minimal sufficient for $\theta$. As for completeness of $X$, notice that $$E_{\theta}[\sin (2\pi X)]=\int_{\theta}^{\theta+1}\sin (2\pi x)\,dx=0\quad,\,\forall\,\theta\in\mathbb R$$
However $\sin (2\pi X)$ is not almost surely $0$, so that $X$ is not a complete statistic.
In fact a complete sufficient statistic does not exist for this model.
To see whether UMVUE of $g(\theta)$ actually exists or not, recall the necessary-sufficient condition for an unbiased estimator (with finite second moment) to be the UMVUE which says that the unbiased estimator has to be uncorrelated with every unbiased estimator of zero.
If possible, suppose $T$ is UMVUE of $g(\theta)$. Let $\mathcal U_0$ be the class of all unbiased estimators of zero.
Clearly for every $H\in \mathcal U_0$,
$$\int_{\theta}^{\theta+1}H(x)\,dx=0\quad,\,\forall\,\theta\in\mathbb R$$
Differentiating both sides of the last equation with respect to $\theta$ gives
$$H(\theta+1)=H(\theta)\quad,\,\text{a.e.}\tag{1}$$
As $T$ is UMVUE, $E_{\theta}(TH)=0$ for all $\theta$ and for all $H\in \mathcal U_0$. In other words, $TH\in \mathcal U_0$ whenever $H\in \mathcal U_0$. So analogous to $(1)$ we have
$$T(\theta+1)H(\theta+1)=T(\theta)H(\theta)\quad,\,\text{a.e.}\tag{2}$$
And $(1)$ implies $$T(\theta)=T(\theta+1)\quad,\,\text{a.e.}\tag{3}$$
Again as $T$ is unbiased for $\theta$, $$\int_{\theta}^{\theta+1} T(x)\,dx=g(\theta)\quad,\,\forall\,\theta\in\mathbb R $$
Differentiating both sides wrt $\theta$ and equation $(3)$ yields
$$g'(\theta)=T(\theta+1)-T(\theta)=0\quad,\,\text{a.e.}$$
This shows that $g(\theta)$ does not admit a UMVUE for any non-constant $g$.
So if you take $g(\theta)=\theta$, then $T=X-\frac12$ is unbiased for $\theta$ but $T$ is not UMVUE.
As for the second question, even if $T$ is just an unbiased estimator (efficient or not) of $\theta$, it does not mean $g(T)$ is unbiased (forget UMVUE) for $g(\theta)$ for an arbitrary nonlinear function $g$.
Among several possible examples, consider i.i.d observations $X_1,\ldots,X_n$ having an exponential distribution with mean $\theta$. Then it is easy to verify that the sample mean $\overline X$ is an efficient estimator (and UMVUE) of $\theta$ but $\overline X^2$ is not UMVUE of $\theta^2$.
Best Answer
For part c) expanding on the hint, fix one of the observations $X_i$ (any one of the $n$ will do). WLOG, let it be $X_1$ and let an estimator be
$$W= \begin{cases} 1 & \text{if }X_1 \geq 1 \\ \\ 0 & \text{if }X_1 = 0 \end{cases} $$ Then $EW = P(X_1 \geq 1) = 1- P(X_1 =0) = 1-\exp(-\theta) = \eta$. So $W$ is an unbiased estimator (even though not a "sensible" one) of $\eta$. Now do "Rao-Blackwellizing", i.e., find $E(W \, | \, Y = y)$. The result will be UMVUE for $\eta$.