Answer: (a) \(x\lt 1\) or \(x\gt 4\); (b)(i) \(3(x-2)^2+4\); (b)(ii) \(y=4\).

For a quadratic expression, first find its roots or complete the square. The sign of the quadratic then follows from its shape and the position of the roots.

(a) Start with

\(3x^2-12x+16\gt 3x+4.\)

Bring all terms to one side:

\(3x^2-15x+12\gt 0.\)

Factorise:

\(3(x^2-5x+4)\gt 0,\)

so

\(3(x-1)(x-4)\gt 0.\)

The quadratic is positive outside its roots, so

\(x\lt 1\quad\text{or}\quad x\gt 4.\)

(b)(i) Complete the square:

\(3x^2-12x+16=3(x^2-4x)+16.\)

Since

\(x^2-4x=(x-2)^2-4,\)

we get

\(3x^2-12x+16=3\bigl((x-2)^2-4\bigr)+16.\)

Hence

\(3x^2-12x+16=3(x-2)^2+4.\)

(b)(ii) The minimum point is \((2,4)\). At the minimum point, the tangent is horizontal, so its equation is

\(y=4.\)