Is $T_n$ an unbiased estimator of $\theta$? Prove your answer. 
Let $(x_1, x_2, \dots, x_n)$ be an observed sample from a distribution with probability density function given by $$f(x) =\begin{cases}\displaystyle\frac{1}{5\theta+8}&\text{if }0 \leq x \leq 5 \theta + 8\\0&\text{otherwise. }\end{cases}$$
Where $\theta \in \mathbb R^+$ is an unknown parameter. It can be shown that $T_n = \frac{X_{(n)}-8}{5}$ is an $MLE$ of $\theta$, where $X_{(n)}$ is the $n^{th}$ order statistic. $\big($i.e. max$\{X_1, X_2, \dots, X_n\}$$\big)$.
(a) Is $T_n$ an unbiased estimator of $\theta$? Prove your answer.
(b) Is $T_n$ asymptotically unbiased for $\theta$? Give reasons for your answer.
(c) Is $T_n$ consistent for $\theta$ in probability? Prove your answer.


ATTEMPT
$(a)$
$$F_{X}(x) = \int_{0}^{x} f(t)dt = \int_{0}^{x} \frac{1}{5 \theta +8} dt = \frac{1}{5\theta+8}t|_{0}^{x} = \frac{1}{5\theta+8}x$$
$$F_{X_{(n)}}(x) = \left(\frac{x}{5\theta+8}\right)^n$$
$$f_{X_{(n)}}(x) = \frac{dF_{X_{(n)}}(x)}{dx} = \frac{n}{(5 \theta + 8)^n}x^{n-1}, 0 \leq x \leq 5 \theta + 8$$
$$\mathsf{E}T_n = \mathsf{E}\left[\left(\frac{X_{(n)} - 8}{5}\right)\right] = 1/5(\mathsf{E}(X_{(n)}) - 8)$$
$$\begin{align}\mathsf{E}(X_{(n)}) = \int_{0}^{5 \theta + 8} x f_{X_{(n)}}(x)dx &= \int_{0}^{5 \theta + 8}\frac{n}{(5 \theta + 8)^n}x^{n}dx \\ &= \frac{n}{(5 \theta + 8)^n(n+1)}x^{n+1}\bigg|_{0}^{5 \theta+8} = \frac{n(5 \theta + 8)^{n+1}}{(5 \theta + 8)^n(n+1)} = \frac{(5 \theta + 8)n}{n+1}\end{align}$$
$$\mathsf{E}T_n = 1/5(\mathsf{E}(X_{(n)}) - 8) = 1/5 \left (\frac{(5 \theta + 8)n}{n+1} - 8 \right) \neq \theta,$$  so $T_n$ is not an unbiased estimator of $\theta.$
$(b)$ Yes, because $\lim\limits_{n\to\infty} \mathsf{E}T_n = \lim\limits_{n\to\infty} 1/5 \left (\dfrac{(5 \theta + 8)n}{n+1} - 8 \right) = \theta$
$(c)$ $$\begin{align}\mathsf{E}(X^2_{(n)}) = \int_{0}^{5 \theta + 8} x^2 f_{X_{(n)}}(x)dx &= \int_{0}^{5 \theta + 8}\frac{n}{(5 \theta + 8)^n}x^{n+1}dx \\ & = \frac{n}{(5 \theta + 8)^n(n+2)}x^{n+2}\bigg|_{0}^{5 \theta+8} = \frac{n(5 \theta + 8)^{n+2}}{(5 \theta + 8)^n(n+2)}= \frac{(5 \theta + 8)^2n}{n+2}\end{align}$$
$$\begin{align}\lim\limits_{n\to\infty} \mathsf{Var}(X_{(n)}) = \lim\limits_{n\to\infty} \left(\mathsf{E}X^2_{(n)} - (\mathsf{E}X_{(n)})^2\right) = \lim\limits_{n\to\infty} \left(\frac{(5 \theta + 8)^2n}{n+2} -  \left(\frac{(5 \theta + 8)n}{n+1}\right)^2\right) = 0\end{align}$$
$T_n$ is consistent for $\theta$ in probability because $(1)$ $T_n$ is asymptotically unbiased for $\theta$ and $(2)$ $\lim\limits_{n\to\infty} \mathsf{Var}(X_{(n)}) = 0$

Is this correct?

The computation in (a) is not strictly necessary, because $$\Pr[T_n > \theta] = 0,$$ whereas $\Pr[T_n < \theta] > 0$; therefore, $\operatorname{E}[T_n] < \theta$ hence is trivially biased.
However, once the computation is performed, (b) immediately follows, and (c) is also correct.
