答案：
【解】(1) 由 $E(X)=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x \mathrm{~d} x=\left.\frac{1}{2} x^2\right|_{\theta-\frac{1}{2}} ^{\theta+\frac{1}{2}}=\theta, \bar{X}=\frac{1}{n} \sum_{i=1}^n X_i$, 令 $E(X)=\bar{X}$, 得 $\hat{\theta}_M=\bar{X}$
设 $x_1, x_2, \cdots, x_n$ 为简单随机样本 $X_1, X_2, \cdots, X_n$ 的样本值, 则似然函数为
$$
\begin{aligned}
L(\theta) & =\prod_{i=1}^n f\left(x_i ; \theta\right)= \begin{cases}1, & \theta-\frac{1}{2} \leqslant x_1, x_2, \cdots, x_n \leqslant \theta+\frac{1}{2}, \\
0, \quad \text { 其他 }\end{cases} \\
& = \begin{cases}1, & \theta-\frac{1}{2} \leqslant \min \left\{x_1, x_2, \cdots, x_n\right\} \leqslant x_1, x_2, \cdots, x_n \leqslant \max \left\{x_1, x_2, \cdots, x_n\right\} \leqslant \theta+\frac{1}{2}, \\
0, & \text { 其他, }\end{cases}
\end{aligned}
$$
由最大似然估计量定义,
$$
\left\{\begin{array}{l}
\hat{\theta}_L-\frac{1}{2} \leqslant X_{(1)}, \\
\hat{\theta}_L+\frac{1}{2} \geqslant X_{(n)},
\end{array}\right.
$$
故满足 $X_{(n)}-\frac{1}{2} \leqslant \hat{\theta}_L \leqslant X_{(1)}+\frac{1}{2}$ 的统计量均为 $\theta$ 的最大似然估计量.
(2) $\frac{X_{(1)}+X_{(n)}}{2}$ 为区间 $\left[X_{(n)}-\frac{1}{2}, X_{(1)}+\frac{1}{2}\right]$ 的中点, 即其为 $\theta$ 的一个最大似然估计量, 由
$$
F(x ; \theta)= \begin{cases}0, & x < \theta-\frac{1}{2}, \\ x-\theta+\frac{1}{2}, & \theta-\frac{1}{2} \leqslant x < \theta+\frac{1}{2}, \\ 1, & x \geqslant \theta+\frac{1}{2},\end{cases}
$$
有
$$
\begin{gathered}
f_{(1)}(x)=n[1-F(x ; \theta)]^{n-1} f(x ; \theta)= \begin{cases}n\left(\frac{1}{2}+\theta-x\right)^{n-1}, & \theta-\frac{1}{2} \leqslant x \leqslant \theta+\frac{1}{2}, \\
0, & \text { 其他, }\end{cases} \\
f_{(n)}(x)=n[F(x ; \theta)]^{n-1} f(x ; \theta)= \begin{cases}n\left(x+\frac{1}{2}-\theta\right)^{n-1}, & \theta-\frac{1}{2} \leqslant x \leqslant \theta+\frac{1}{2}, \\
0, & \text { 其他, }\end{cases}
\end{gathered}
$$
$$
\begin{aligned}
\text { 故 } E\left[X_{(1)}\right] & =\int_{-\infty}^{+\infty} x f_{(1)}(x) \mathrm{d} x=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x n\left(\frac{1}{2}+\theta-x\right)^{n-1} \mathrm{~d} x \stackrel{\frac{1}{2}+\theta-x=t}{=} \int_0^1\left(\frac{1}{2}+\theta-t\right) t^{n-1} \mathrm{~d} t \\
& =\theta+\frac{1}{2}-\frac{n}{n+1}, \\
& E\left[X_{(n)}\right]=\int_{-\infty}^{+\infty} x f_{(n)}(x) \mathrm{d} x=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x n\left(x+\frac{1}{2}-\theta\right)^{n-1} \mathrm{~d} x=\theta-\frac{1}{2}+\frac{n}{n+1},
\end{aligned}
$$
故
$$
E\left[\frac{X_{(1)}+X_{(n)}}{2}\right]=\frac{1}{2}\left\{E\left[X_{(1)}\right]+E\left[X_{(n)}\right]\right\}=\frac{1}{2}\left(\theta+\frac{1}{2}-\frac{n}{n+1}+\theta-\frac{1}{2}+\frac{n}{n+1}\right)=\theta,
$$
所以 $\frac{X_{(1)}+X_{(n)}}{2}$ 为 $\theta$ 的无偏估计.