Number of iterations of the Euclidean algorithm
There's very little to deduce.

The number of iterations is linear in the size of the input because the size of the input is $2+\log_2 a + \log_2 b$ and the number of iterations is $2\log_2 a$, which is less than $2(2+\log_2 a + \log_2 b)$, so is less than twice the length of the input.  Hence, (2) holds.

Each iteration requires only a polynomial number of steps: all it does is tests and simple arithmetic operations, all of which can be done in time polynomial in the number of bits involved.  Therefore, the total required time is some linear function (the number of iterations) times some polynomial function (the cost of each iteration), which is another polynomial function. So (3) holds.