Answer: \(a=3\), \(b=8\), \(c=4\).

The amplitude is \(|a|\). Since the amplitude is \(3\), take

\(a=3.\)

The period of \(\cos(bx)\) is

\(\dfrac{2\pi}{b}.\)

So

\(\dfrac{2\pi}{b}=\dfrac{\pi}{4}.\)

Hence

\(b=8.\)

Now use the point \(\left(\dfrac{\pi}{12},\dfrac52\right)\):

\(\dfrac52=3\cos\left(8\cdot\dfrac{\pi}{12}\right)+c.\)

Since \(8\cdot\dfrac{\pi}{12}=\dfrac{2\pi}{3}\) and \(\cos\dfrac{2\pi}{3}=-\dfrac12\),

\(\dfrac52=3\left(-\dfrac12\right)+c.\)

Therefore

\(\dfrac52=-\dfrac32+c,\)

so

\(c=4.\)

Thus

\(\boxed{a=3,\quad b=8,\quad c=4}.\)