Answer: \(a=4\), \(b=6\), \(c=-2\).

The amplitude is \(|a|\). Since the amplitude is \(4\), take

\(a=4.\)

The period of \(\sin(bx)\) is \(\dfrac{2\pi}{b}\). Therefore

\(\dfrac{2\pi}{b}=\dfrac{\pi}{3}.\)

So

\(b=6.\)

Use the point \(\left(\dfrac{\pi}{12},2\right)\):

\(2=4\sin\left(6\cdot\dfrac{\pi}{12}\right)+c.\)

Since \(6\cdot\dfrac{\pi}{12}=\dfrac{\pi}{2}\) and \(\sin\dfrac{\pi}{2}=1\),

\(2=4+c.\)

Hence

\(c=-2.\)

Therefore

\(\boxed{a=4,\quad b=6,\quad c=-2}.\)