Let $X_1,..,X_n$ be For a random sample $X_1,…,X_n$ from a Uniform$[\theta,θ+1] $distribution, with probability density function $f(x;\theta) = \begin{cases} 1, & \theta \le x \le \theta+1 \\ 0, & \text{otherwise} \end{cases}$ and $X_{(n)}=max(X_1,X_2,…,X_n)$ and $X_{(1)}=min(X_1,...,X_n)$. I want to show:
1)$X_{(1)}$ has the same distribution as $\theta - X_{(n)}$
2)$\hat{\theta}=X_{(1)}+X_{(n)}$ is an unbiased estimator of θ.
My work for 1) I know that $X_{(1)}$ has cdf $F_{X_{(1)}}(x)=1-(1-F(x))^n=\begin{cases} 0, & x\le\theta \\ 1-(1-(x-\theta))^n, & \theta< x<\theta+1 \\ 1, &\theta+1 \le x \end{cases}$.
The numbers seem to be a little off. You can say, however, that $X_{(1)}$ is distributed like $2\theta + 1 - X_{(n)}$, while $\frac{1}{2}(X_{(1)} + X_{(n)} - 1)$ is an unbiased estimator of $\theta$ (as @Henry pointed out in the comments).
The first problem can be solved using a direct computation, but there's no need to expand any expressions or do any complicated algebra.
$$F_{X_{(1)}}(x) = 1 - (1 - F(x))^n.$$
as you noted. We can also compute,
$$F_{X_{(n)}}(x) = P(X_{(n)} < x) = P(X_k < x\text{ for } k=1,\dots,n) = (F(x))^n$$
Then,
$$F_{2\theta+1-X_{(n)}}(x) = P(2\theta + 1 - X_{(n)} < x) = P(X_{(n)} > 2\theta + 1 -x) = 1 - (F(2\theta+1-x))^n$$
However, we know that $F(x) = x - \theta$ for $\theta \leq x \leq \theta+1$, so
$$F(2\theta + 1 - x) = \theta + 1- x = 1 - (x - \theta) = 1 - F(x)$$
Conclude,
$$F_{2\theta+1-X_{(n)}}(x) = 1 - (F(2\theta+1-x))^n = 1 - (1 - F(x))^n = F_{X_{(1)}}(x).$$
For the second problem, just apply the first:
$$E[X_{(1)} + X_{(n)}] = E[2\theta + 1 - X_{(n)}] + E[X_{(n)}] = 2\theta + 1.$$