It's false as you state it. Instead of the condition $X'_j\perp X_j$, you need that $X'_j\perp(X_i, i\ne j)$, so that $X_1, X_2, \dots, X'_j, \dots, X_n$ is also an i.i.d. collection.
Under that condition, the vectors
$(X_1, \dots, X_j, \dots, X_n)$ and
$(X_1, \dots, X'_j, \dots, X_n)$
have the same distribution. So if $g$ is any function, then
$W:=g(X_1, \dots, X_j, \dots, X_n)$ and $W':=g(X_1, \dots, X'_j, \dots, X_n)$ have the same distribution.
Now if you've got some further thing that "does not depend on the $j$th variable", say $Y=h(X_1, \dots, X_{j-1}, X_{j+1}, \dots, X_n)$, then
$W-Y$ and $W'-Y$ have the same distribution. This is just another application of the previous paragraph, since $W-Y$ is a function of $X_1,\dots,X_j,\dots, X_n$
and $W'-Y$ is "the same function" of $X_1,\dots, X'_j,\dots, X_n$.
However, if you had a second quantity that did depend on the $j$th coordinate, say
$Z=r(X_1, \dots, X_j, \dots, X_n)$, then you couldn't conclude that $W-Z$ and $W'-Z$ have the same distribution. In that case $W'-Z$ would be a function of all $n+1$ variables
$X_1,\dots,X_j, X'_j, \dots, X_n$.
Best Answer
We want the probability that at least $500$ of the $X_i$ are greater than $0.01$.
There are various ways to do this. The probability that an individual $X_i$ is greater than $0.01$ is $0.49$. Call an $X_i\gt 0.01$ a success. We want the probability that the number of successes in $999$ trials is $\ge 500$. A standard approach to this is the normal approximation to the binomial, with continuity correction.
For the probability that the empirical median is $X_1$, the empirical median is equally likely to be any of the observations.