(a) We have \(f'(x) = \frac{-3}{(x+2)^4}\) and the gradient of the tangent is \(-\frac{16}{27}\) at \(x = a\). Set \(\frac{-3}{(a+2)^4} = -\frac{16}{27}\).

Equating gives \(16(a+2)^4 = 81\).

Solving for \((a+2)^2\), we get \((a+2)^2 = \frac{9}{4}\).

Thus, \(a+2 = \pm \frac{3}{2}\).

Solving for \(a\), we find \(a = \frac{1}{2}\) or \(a = -\frac{7}{2}\).

(b) To find \(f(x)\), integrate \(f'(x) = \frac{-3}{(x+2)^4}\).

\(f(x) = \int \frac{-3}{(x+2)^4} \, dx = \frac{-1}{(x+2)^3} + c\).

Given \(f(-1) = 5\), substitute to find \(c\):

\(5 = \frac{-1}{1^3} + c \Rightarrow 5 = -1 + c \Rightarrow c = 6\).

Thus, \(f(x) = \frac{-1}{(x+2)^3} + 6\).