For $\operatorname{U}(\theta-1,\theta+1)$ the mean, median and centre of symmetry all coincide at $\theta$ , and the variance is $\dfrac{(2)^{2}}{12}=\dfrac13$ .

(B) is true. $\mathbb{E}(\overline{X})=\theta$ and

$\operatorname{Var}(\overline{X})=\frac{1/3}{2n+1}\longrightarrow0 ,$

so by Chebyshev $\overline{X}\overset{P}{\longrightarrow}\theta$ : the sample mean is consistent.

The others fail:

(A) is wrong in its second half. The sample median converges in probability to the population median, which here is $\theta$ . So $T_2$ is consistent too, and (B) — which claims nothing about $T_2$ — is the safe statement.
(C) is wrong. The distribution is symmetric about $\theta$ , so the sample median is also unbiased, and hence so is $T_3=(T_1+T_2)/2$ . "Only $T_1$ " is false.
(D) is wrong. For a uniform shift family the sufficient statistic is the pair of extremes $\bigl(X_{(1)},X_{(2n+1)}\bigr)$ , since the likelihood is $2^{-(2n+1)}$ on $\{\,x_{(2n+1)}-1<\theta<x_{(1)}+1\,\}$ and zero elsewhere. The median carries no information about where that interval sits.