- This is, of course, the same as saying about spacings between uniform points on a segment (you can say that $Y_1=0$, for example). Let it be the segment $[0,1]$.
Now the joint distribution of $I_1,\dots, I_{n}$ is the same as of $E_1/E,\dots, E_n/E$, where $E_1,\dots, E_n$ are iid exponential distributed, $E=\sum_{k=1}^n E_k$ (see Devroye Non-Uniform Random Variate Generation, p.208). So the distribution of $I_{(1)},\dots, I_{(n)}$ is the same as of $E_{(1)}/E,\dots, E_{(n)}/E$. But the joint distribution of $\{E_{(k)}-E_{(k-1)},k=1,\dots,n\}$ ($E_{(0)}:=0$) is the same as of $\{(n-k+1)^{-1} E_k,k=1,\dots,n\}$ (ibid, p.211).
So the distribution of $J_1,\dots, J_n$ is the same as of $\{(n+k-1)^{-1} E_k/E,k=1,\dots,n\}$, where $E_1,\dots, E_k$ are iid exponential rv's, $E=\sum_{k=1}^n E_k$. And this is, by the previous paragraph, equivalent to saying that the distribution is the same as of $\{(n-k+1)^{-1} I_k,k=1,\dots,n\}$.
These are not independent, but very close to be, and from here you can find the distribution of maximum and minimum (but nothing very pleasant there, as the variables in question are not identically distributed; a formula for the expectation looks extremely ugly).
How to get distribution of $J$ omitting $E$. In fact, this is simple owing to the fact that the ordering map on the simplex $\{(t_1,\dots,t_n)|t_j\ge 0,\sum_j t_j=1\}$ (the support of $I$) is picewise linear, and moreover each image has the same number of preimages due to the apparent symmetry. So the distribution of $\{I_{(1)},\dots,I_{(n)}\}$ is uniform on its support. Now we have a one-to-one linear map to $J$. So $J$ is also uniformely distributed. So it's only about finding its support, which is simple, as John Jiang noted.