My Question：
$X_1,X_2,...X_n$ are iid random varables and follow Uniform$(0,\theta)$ We pick up five samples from population.And Let $Y_1<Y_2<Y_3<Y_4<Y_5$. Want to know $E(2Y_2|Y_5)$
My Attempt：
$fY_2,Y_5(s,t)$
=$\frac{5!}{2!}(\frac{t}{\theta}-\frac{s}{\theta})^2(\frac{s}{\theta})(\frac{1}{\theta})$
=$\frac{5!}{2!}\frac{1}{\theta^5}(t-s)^2s$
$0<s<t<\theta$
$fY_5(t)$
=$\int_{0}^t\frac{5!}{2!}\frac{1}{\theta}(t-s)^2s$ds
=$\frac{5}{\theta^5}t^2$
$fY_2|Y_5(s|t)$ =$\frac{\frac{5!}{2!}\frac{1}{\theta^5}(t-s)^2s}{\frac{5}{\theta^5}t^2}$ =$\frac{6(t-s)^2s}{t^2}$
$EY_2|Y_5(s|t)$ =$\int_{0}^t \frac{6(t-s)^2s}{t^2}$ds
I think It is wrong,because $\theta$ is disappear in $fY_2|Y_5(s|t)$. I find that $Y_5$ is sufficient and complete statistic.How to use this condition to get the answer?
Your approach is correct but your computations contain errors.
First, your joint density for $(Y_2, Y_5)$ is correct. However, your marginal density is not correct:
$$\begin{align} f_{Y_5}(t) &= \frac{60}{\theta^5} \int_{s=0}^t s (s-t)^2 \, ds \\ &= \frac{60}{\theta^5} s^3 - 2ts^2 + t^2 s \, ds \\ &= \frac{60}{\theta^5} \left[ \frac{s^4}{4} - \frac{2ts^3}{3} + \frac{t^2 s^2}{2} \right]_{s=0}^t \\ &= \frac{60}{\theta^5} \left(\frac{t^4}{4} - \frac{2t^4}{3} + \frac{t^4}{2} \right) \\ &= \frac{5 t^\color{red}{4}}{\theta^5}, \quad 0 < t < \theta. \tag{1} \end{align}$$
Then the conditional density of $Y_2 \mid Y_5$ is $$f_{Y_2 \mid Y_5}(s \mid t) = \frac{f_{Y_2, Y_5}(s,t)}{f_{Y_5}(t)} = \frac{\frac{60}{\theta^5} s(s-t)^2}{\frac{5t^4}{\theta^5}} = \frac{12s(s-t)^2}{t^4}, \quad 0 < s < t. \tag{2}$$
The conditional expectation is $$\operatorname{E}[2Y_2 \mid Y_5 = t] = \int_{s=0}^t (2s) f_{Y_2 \mid Y_5} (s \mid t) \, ds = \frac{4t}{5}. \tag{3}$$
Indeed, the answer is not a function of $\theta$. Intuitively, the reason is because once you know $Y_5$, you also know that $Y_2 < Y_5$, and $\theta$ no longer matters. If I said $Y_5 = 10$, then it is irrelevant whether $\theta = 11$ or $\theta = 10000$. Equation $(2)$ tells you that the conditional distribution of $Y_2$ given $Y_5$ is not a function of $\theta$, so the expectation is also independent of $\theta$. Our answer, $4t/5$, also makes intuitive sense: if the sample is uniformly distributed on $[0,\theta]$, we would expect that $Y_2$ is about $2/5$ the value of $Y_5$, so on average, $2Y_2$ should be $4/5$ of $Y_5$.