Copies sit together

In a sorted list, all copies of a value sit in one block, like every copy of the same book on a library shelf. So counting them is "where does the block end, minus where does it start".

You already have a search for the start: the first position holding a value at least x.

Start of the block: the first position with a value at least x.

How many 5s?

startend

Search 1: the first value ≥ 5 is at position 1. The 5s start here.

Move 1 of 3

The end is another start

Just past the last x is the first value bigger than x. For whole numbers, that is the first value at least x plus one. So run the same search again with x plus one.

Put the search in its own function and call it twice. The difference of the two answers is the count.

Just past the block: the first position with a value at least x plus one.

x isn't there? Both searches land on the same spot. The count is 0, with no special case.

In code

def first_at_least(x):
    low, high = 0, len(nums)
    while low < high:
        mid = (low + high) // 2
        if nums[mid] >= x:
            high = mid
        else:
            low = mid + 1
    return low

answers = []
for x in queries:
    start = first_at_least(x)
    end = first_at_least(x + 1)
    answers.append(end - start)
return answers