Straightforward

At first, replace the given bracket sequence by an integer array $a$.

In this array,

a[i] = count of opening bracket till index i - count of closing bracket till index i

Now in each query, we need to find the maximum possible index x, in the range from l to r where $a[x]=a[l-1]$ and no element from $a[l]$ to $a[x]$ is less than $a[l-1]$. To do this, first check if there is any element from $a[l]$ to $a[r]$ that is less than $a[l-1]$. Find the first occurrence of such an element. This can be done using segment tree in $O(N \times log(N))$ time complexity. If such an element exists, replace $r$ with the index of that element. Then find the maximum index $y$ between $l$ and $r$ where $a[y]=a[l-1]$. This can be found by using binary search. The output will be $y-l+1$.

Time complexity $O(N × log(N))$.

This problem can also be solved by using sparse table. Time complexity is same $O(N × log(N))$.