(i) To find the largest value of \(a\) for which the composite function \(gf\) can be formed, we need \(3x + 1 \leq -1\). Solving for \(x\), we have:

\(3x + 1 \leq -1\)

\(3x \leq -2\)

\(x \leq -\frac{2}{3}\)

Thus, the largest value of \(a\) is \(-\frac{2}{3}\).

(ii) For \(a = -1\), solve \(fg(x) + 14 = 0\):

\(fg(x) = 3(-1 - x^2) + 1\)

\(fg(x) + 14 = 0 \Rightarrow 3(-1 - x^2) + 1 + 14 = 0\)

\(3(-1 - x^2) + 15 = 0 \Rightarrow 3(-1 - x^2) = -15\)

\(-3 - 3x^2 = -15 \Rightarrow 3x^2 = 12\)

\(x^2 = 4 \Rightarrow x = -2\) (since \(x \leq -1\))

(iii) Find the set of values of \(x\) which satisfy \(gf(x) \leq -50\):

\(gf(x) = -1 - (3x + 1)^2\)

\(gf(x) \leq -50 \Rightarrow -1 - (3x + 1)^2 \leq -50\)

\(-(3x + 1)^2 \leq -49 \Rightarrow (3x + 1)^2 \geq 49\)

\(3x + 1 \geq 7\) or \(3x + 1 \leq -7\)

\(3x \geq 6 \Rightarrow x \geq 2\) or \(3x \leq -8 \Rightarrow x \leq -\frac{8}{3}\)

Since \(x \leq -1\), the solution is \(x \leq -\frac{8}{3}\).