You managed to find the first part, that is the unbiased estimator for $\theta$, by noting:
$$E[X] = \theta + \frac{1}{2} \implies \hat{\theta} = \bar{X} - \frac{1}{2}$$
Now you want a confidence interval at the $\beta$ level.
This means that you want $$P_{\bar{X}}[a\leq\bar{X}-\frac{1}{2} \leq b] = 1-\beta$$
So what you really need is to find the distribution of $\bar{X}$.
Now you can find that $$t = \frac{\bar{X} - \mu}{(\frac{s}{\sqrt{n}})}$$
Where $t$ follows a $t-distribution$ with $n-1$ degrees of freedom. Where $s$ is your sample standard deviation (coming from your second moment).
Generally when you find a confidence interval you want it to be of minimum measure/cardinality/length, however in this case the distribution is symmetric, so you don't need to struggle to find the range as it is symmetric around your estimator.
I encourage you to find out why the above t distribution is true!
Hope this helps.
______EDIT_____
As noted by jbowman in the comments, we already know the standard deviation, in which case we can use :
$$ z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$$
Where $\sigma$ is the "known" standard deviation and $z$ follows a standard normal distribution
The result is correct, but the reasoning is somewhat inaccurate. You need to keep track of the property that the density is zero outside $[0,\theta]$. This implies that the likelihood is zero to the left of the sample maximum, and jumps to $\theta^n$ in the maximum. It indeed decreases afterwards, so that the maximum is the MLE.
This also entails that the likelihood is not differentiable in this point, so that finding the MLE via the "canonical" route of the score function is not the way to go here.
A more detailed formal derivation is, e.g., given here
Best Answer
Consider a similar problem with a specific data set
$X\sim U(0,\beta)$ (continuous uniform)
Let $x_1=4.31$, $x_2=1.24$, $x_3=5.15$
Note that $0<X_i<\beta$ and so in turn $0<x_i<\beta$.
Consequently, $\beta<x_i$ for any $x_i$ is not a possible value for the parameter.
As a result the likelihood function for this slightly different problem looks like this:
Now consider a discrete (integer-valued) uniform $U[0,a]$ and the observations 4, 1 and 5. Can you draw the likelihood (hint: don't draw a curve again)
Then tackle the original problem. Make sure your answer makes sense for the $x_1=1,x_2=-1000$ case Alex R mentioned in comments.