a)
[tex]f(x) = \sqrt{x^{2}-16}\\\\\sqrt{x^{2}-16} \geq 0\\\\x^{2}-16 \geq 0\\\\(x+4)(x-4) = 0\\\\x = -4 \ \vee \ x = 4\\\\a > 0\\\\\boxed{D: \ x\in(-\infty; -4\rangle \ \cup \ \langle4;+\infty)}[/tex]
b)
[tex]f(x) = \sqrt{x^{2}-4x+3}\\\\\sqrt{x^{2}-4x+3} \geq 0\\\\x^{2}-4x+3 \geq 0\\\\M. \ zerowe:\\\\x^{2}-4x+3 = 0\\\\a = 1, \ b = -4, \ c = 3\\\\\Delta = b^{2}-4ac = (-4)^{2}-4\cdot1\cdot3 = 16-12 = 4\\\\\sqrt{\Delta } = \sqrt{4} = 2\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{4-2}{2\cdot1} = \frac{2}{2} = 1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4+2}{2} = \frac{6}{2} = 3\\\\a > 0\\\\\boxed{D: \ x\in(-\infty;1\rangle \ \cup \ \langle3;+\infty)}[/tex]
