Yes, this is solvable: the queries are equivalent to "is $$$[l, r - 1]$$$ in the difference array all-positive" which can be solved using a segment tree. The point updates don't break this solution either: a point update only changes 2 elements of the difference array so you can do 2 point updates in the constructed segment tree.

I think we can solve it using Fenwick tree

Here is my solution -->

Traverse the array and check if(A[i]>A[i-1]) then update value for i to be 1 using update operation other wise keep it 0.

for the subarray(l,r) to be increasing sum(r) — sum(l) == r-l; (NOTE: we have take l not l-1)

for point update just check it with its neighbouring elements and update accordingly.