I will show the lower and upper bound.

First we compute the cost of g3.

Take for example, n = 2^16.

How many iterations we make in the for loop?

i=2^0, i=2^1, i=2^2, i=2^3... < 2^16

More or less, that would be 16 steps. So the cost of g3 is O(log(n)).

Now lets try to compute f3. Since it's using g3 inside the loop, it would go as follows:

log(n) + log(n^(1/2)) + log(n^(1/4)) + log(n^(1/8)) + ...

That's for sure greater than log(n), so we could take log(n) as the lower bound.

Now, in order to compute the upper bound we have to think, how many iterations does the loop do?

Take again 2^16 as an example:

2^16, 2^16^(1/2), 2^16^(1/4), 2^16^(1/8), 2^16^(1/16),

That turns out to be:

2^16, 2^8, 2^4, 2^2, 2^1

And in the next iteration we would stop because sqrt(2) rounds to 1.

So in general, if n=2^2^k, we make k iterations. That's log(log(n)). That means we could say log(n)*log(log(n)) as the upper bound.

There is probably a more adjusted solution but this should be pretty accurate.