Can we create faster sort algorithm than O(N log N)
You got a result of $O(N\cdot \log(\sqrt N)+N\log(\sqrt N))$.

But $\log \sqrt N = (\log N)/2$, so $N⋅log(\sqrt N)+N·\log(\sqrt N) = N·\log N$. So not only is it the same complexity class, where constant factors don't matter, it is even the same function, therefore the same constant factor. 

But the underlying reason for any comparison based sorting algorithm to take $O(\log N)$ comparisons is that there are N! possible permutations of the array elements, each comparison can rule out at most half the possible cases, so you absolutely need at least $\log_2 N!$ comparisons in the worst case, which is $O(N\cdot  \log N)$. That $\log_2 N!$ is not a big-O, it is an absolute number. You cannot achieve for example $0.999 \log_2 N!$ comparisons in the worst case. 

So you would have to come up with an algorithm that isn't based on comparison. For example, if an image contains pixels using 24 bits each, then you can sort the $N$ pixels of an image in $O(N)$ (see "The Player of Games" for reasons why you would want to sort all the pixels in a large library).