You have a $U(-\theta,\theta)$ population where $\theta\in\mathbb R^+$.
Joint density of the sample $\mathbf X=(X_1,X_2,\ldots,X_n)$ is
\begin{align}
f_{\theta}(\mathbf x)&=\frac{1}{(2\theta)^n}\mathbf1_{-\theta < x_1, \ldots, x_n < \theta}
\\&=\frac{1}{(2\theta)^n}\mathbf1_{0<|x_1|,\ldots,|x_n|<\theta}
\\&=\frac{1}{(2\theta)^n}\mathbf1_{\max_{1\le i\le n}|x_i|<\theta}
\end{align}
It is clear from Factorization theorem that a sufficient statistic for $\theta$ is $$T(\mathbf X)=\max_{1\le i\le n}|X_i|$$
One could verify that $|X_i|\sim U(0,\theta)$, so that the density of $T$ is $$g_{\theta}(t)=\frac{n}{\theta^n}t^{n-1}\mathbf1_{0<t<\theta}$$
That $T$ is a complete statistic for $\theta$ is well-known.
We simply have to find unbiased estimators of the parametric functions of $\theta$ based on the complete sufficient statistic. This would give us the UMVUE by the Lehmann-Scheffe theorem.
As the support of the complete sufficient statistic here depends on the parameter $\theta$, unbiased estimators can be directly obtained through differentiation.
Let $h_1(T)$ and $h_2(T)$ be unbiased estimators of $\theta/(1+\theta)$ and $e^{\theta}/\theta$ respectively, based on the complete sufficient statistic $T$.
That is, for all $\theta>0$,
\begin{align}
\qquad\quad\frac{n}{\theta^n}\int_0^{\theta}h_1(t)t^{n-1}\,dt&=\frac{\theta}{1+\theta}
\\\implies \int_0^{\theta}h_1(t)t^{n-1}\,dt &= \frac{\theta^{n+1}}{n(1+\theta)}
\end{align}
Differentiating both sides wrt $\theta$,
\begin{align}
h_1(\theta)\theta^{n-1}&=\frac{\theta^n(n\theta+n+1)}{n(1+\theta)^2}
\\\implies h_1(\theta) &=\frac{\theta(n\theta+n+1)}{n(1+\theta)^2}
\end{align}
Hence, $$h_1(T)=\frac{T(nT+n+1)}{n(1+T)^2}$$
Similarly for the second problem, for all $\theta>0$,
\begin{align}
\qquad\quad\frac{n}{\theta^n}\int_0^{\theta}h_2(t)t^{n-1}\,dt&=\frac{e^\theta}{\theta}
\\\implies \int_0^{\theta}h_2(t)t^{n-1}\,dt &= \frac{\theta^{n-1} e^\theta}{n}
\end{align}
Differentiating both sides wrt $\theta$ yields
\begin{align}
h_2(\theta)\theta^{n-1}&=\frac{e^{\theta}\theta^{n-2}(\theta+n-1)}{n}
\\\implies h_2(\theta) &=\frac{e^{\theta}(\theta+n-1)}{n\theta}
\end{align}
So, $$h_2(T)=\frac{e^{T}(T+n-1)}{nT}$$
In my initial answer, the following calculation for the UMVUE was rather unnecessary and complicated. Had the support not depended on the parameter, I might have tried this. I am keeping this part in the answer as I might be able to salvage the somewhat faulty argument on some further consideration :
For $k> -n$, we have
\begin{align}
E_\theta(T^k)&=\frac{n}{\theta^n}\int_0^\theta t^{k+n-1}\,dt\\[8pt]
& = \frac{n\theta^k}{n+k}
\end{align}
This suggests that an unbiased estimator of $\theta^k$ based on $T$ is $$\left(\frac{n+k}{n}\right)T^k$$
For the first problem, one could write
\begin{align}
\frac{\theta}{1+\theta}&=
\begin{cases}\left(1+\frac{1}{\theta}\right)^{-1}=1-\frac{1}{\theta}+\frac{1}{\theta^2}-\frac{1}{\theta^3}+\cdots&,\text{ if }\theta>1\\\\\theta(1+\theta+\theta^2+\cdots)&,\text{ if }0<\theta<1\end{cases}
\end{align}
For $0<\theta<1$, we have
$$E_{\theta}\left[\left(\frac{n+1}{n}\right)T+\left(\frac{n+2}{n}\right)T^2+\cdots\right]=\theta+\theta^2+\cdots$$
Or, $$E_{\theta}\left[\sum_{k=1}^\infty\left(\frac{n+k}{n}\right)T^k\right]=\frac{\theta}{1+\theta}$$
For $\theta>1$,
$$E_{\theta}\left[1-\left(\frac{n-1}{n}\right)\frac{1}{T}+\left(\frac{n-2}{n}\right)\frac{1}{T^2}-\cdots\right]=1-\frac{1}{\theta}+\frac{1}{\theta^2}-\cdots$$
That is, $$E_{\theta}\left[\sum_{k=0}^\infty\left(\frac{n-k}{n}\right)\frac{(-1)^k}{T^k}\right]=\frac{\theta}{1+\theta}$$
Hence by Lehmann-Scheffe theorem, UMVUE of $\theta/(1+\theta)$ is
\begin{align}
h_1(T)&=\begin{cases}\displaystyle\sum_{k=1}^\infty\left(\frac{n+k}{n}\right)T^k&,\text{ if }0<\theta<1\\\\\displaystyle\sum_{k=0}^\infty\left(\frac{n-k}{n}\right)\frac{(-1)^k}{T^k}&,\text{ if }\theta\ge1 \end{cases}
\\\\&=\begin{cases}\displaystyle\frac{T(n+1-nT)}{n(T-1)^2}&,\text{ if }0<\theta<1\\\\\displaystyle\frac{T(n+1+nT)}{n(T+1)^2}&,\text{ if }\theta\ge1\end{cases}
\end{align}
However, upon verification of unbiasedness for some values of $n$, it looks like only $$h_1(T)=\displaystyle\frac{T(n+1+nT)}{n(T+1)^2}$$ should be the correct answer for all $\theta>0$. I am not quite sure why that happens.
For the second problem, we can use the power series expansion of $e^\theta$ to obtain
$$E_{\theta}\left[\sum_{k=-1}^{\infty}\left(\frac{n+k}{n}\right)\frac{T^k}{(k+1)!}\right]=\sum_{j=0}^\infty \frac{\theta^{j-1}}{j!}=\frac{e^{\theta}}{\theta}$$
So the UMVUE of $e^{\theta}/\theta$ is
\begin{align}
h_2(T)&=\sum_{k=-1}^{\infty}\left(\frac{n+k}{n}\right)\frac{T^k}{(k+1)!}
\\\\&=\frac{e^T(n-1+T)}{nT}
\end{align}
Best Answer
$\newcommand{\E}{\operatorname{E}}$ \begin{align} \E(-\log X) & = \int_0^1 (-\log x) \Big(\theta x^{\theta-1}\,dx\Big) = \int u\,dv = uv - \int v \, du \\[8pt] & = \left.(-\log x)x^\theta\vphantom{\frac11}\,\right|_0^1 - \int_0^1 x^\theta\Big( \frac{-dx} x \Big) \\[8pt] & = \int_0^1 x^{\theta-1} \,dx \quad(\text{L'Hopital's rule showed that the first term is 0.}) \\[8pt] & = \left.\frac {x^\theta}\theta\right|_0^1 = \frac 1 \theta. \end{align} That takes care of unbiasedness.
The joint density is $$ f(x_1,\ldots,x_n) = \theta^n (x_1\cdots x_n)^{\theta-1} \cdot 1 $$ where the "$1$" is a factor that does not depend on $\theta$ and in this case does not depend on $x_1,\ldots,x_n$, but dependence on those would not upset the following conclusion: the product, and therefore the sum of the logarithms, is sufficient.
You need to show that this sufficient statistic admits no nontrivial unbiased estimators of zero, i.e. there is no nonzero function $g(x_1,\ldots,x_n)$, not depending on $\theta$, for which $$ \int_0^1\cdots\int_0^1 g(x_1,\ldots,x_n)\theta^n(x_1\cdots x_n)^{\theta-1}\,dx_1\cdots dx_n = 0\text{ for all values of $\theta>0$}. $$ (You can divide both sides of that by $\theta^n$ and it's a little bit simpler.)
Maybe I'll be back later to deal with this integral${}\,\ldots\ldots$