You have factored the functions, which tells you where the zeroes are. You'll want to further factor x^2 - 16 ino (x - 4)(x + 4).

Since the functions are continuous, the sign will be constant between the zeroes. All you need to do is check a value in between the zeroes. Alternatively, you can apply what you know about the shape of the function. I'll do the first one to illustrate both approaches.

You know f(x) = x(x-2)(x-4) is a cubic, with zeroes at 0, 2, and 4.

f(-1) = (-1)(-3)(-5) = -15, so it's <= 0 for (-∞,0]

f(1) = 1(-1)(-3) = 3, so it's >0 in (0,2)

f(3) = 3(1)(-1) = -3, so it's <= 0 in [2,4]

f(5) = 5(3)(1) = 15, so it's >0 in (4, ∞)

You also probably know the shape of the graph of a cubic with 3 real zeroes, including the end behaviors. For large negative x, it will be negative, then as you increase the value of x, it becomes 0 at the first zero, then becomes positive, increasing for a while then decreasing until it gets to the 2nd zero., then it becomes negative until you reach the 3rd zero, then it becomes and stays positive. That will give you the same answers for <=0 or > 0 as the above analysis.

You can do the 2nd problem in either of the same ways. Note that the function is 4th degree, and the function will be >0 at both ends, like any other even degree polynomial with a positive coefficient on the largest power term.