$X_1,\cdots,X_5$ is a random sample, $\sim N(0,1)$. Let $T=\sum\limits_{i=1}^5 (X_i-\bar{X})^2$. Find $E(T^2\bar{X}^2)$ $X_1,\cdots,X_5$ is a random sample, $\sim N(0,1)$. Let $T=\sum\limits_{i=1}^5 (X_i-\bar{X})^2$. Find $E(T^2\bar{X}^2)$
My attempt:
$X_i$'s are independent, so $T$ and $\bar{X}$ are independent, so $T^2$ and $\bar{X}^2$ are independent.
$\therefore E(T^2\bar{X}^2)=E(T^2)E(\bar{X}^2)$
Now, $E(\bar{X}^2)=V(\bar{X})+E^2(\bar{X})=V(\bar{X})=\frac15$
How do I find $E(T^2)$?
 A: 
$X_i$'s are independent, so $T$ and $\bar{X}$ are independent

That is not a valid argument at all.
It is true that $T$ and $\overline X$ are independent, but you need to give a valid proof of that. You have
\begin{align}
& \overline X = \frac{X_1+X_2+X_3+X_4+X_5} 5 \\[10pt] \text{and } & T^2 = (X_1-\overline X)^2 + (X_2-\overline X)^2 + (X_3-\overline X)^2 \\
& {} \qquad\qquad {} + (X_4-\overline X)^2 + (X_5-\overline X)^2
\end{align}
Since $\overline X$ and $T^2$ are both defined using the same quintuple, it is hardly obvious that they are independent.
Now note that
$$
\operatorname{cov}(\,\overline X, X_1-\overline X\,) = \frac 1 n - \frac 1 n =0.
$$
That the covariance of two random variables is $0$ falls far short of proving that they are independent, but if $X_1,\ldots,X_n$ are i.i.d. normally distributed, then any two linear combinations
\begin{align}
& a_1 X_1 + \cdots + a_n X_n, \\
& b_1 X_1 + \cdots + b_n X_n,
\end{align}
are independent if their covariance is $0.$ You can use that to show that
$$
\overline X \text{ and } \big(X_1-\overline X,\ldots, X_n-\overline X\big)
$$
are independent, and consequently that $\overline X$ and $T^2$ are independent.
Next you have the question of $\operatorname E(T^2).$
Notice that
\begin{align}
& \operatorname{var}(X_1-\overline X) = \operatorname{var}(X_1) + \operatorname{var}(\,\overline X\,) - 2\operatorname{cov}(X_1,\overline X\,) \\[10pt]
= {} & 1 + \frac 1 n - \frac 2 n = 1 - \frac 1 n.
\end{align}
Since $\operatorname E(X_1-\overline X\,)=0,$ we therefore have
$$
\operatorname E\big( (X_1 - \overline X)^2 \big) = 1 - \frac 1 n = 1 - \frac 1 5.
$$
Now add that up five times.
