Answer: \(a=5\), \(b=74\).

The curve is

\(y=\dfrac{4}{x^2}+ax+7.\)

Differentiate:

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{8}{x^3}+a.\)

At \(x=2\), the gradient of the tangent is

\(-\dfrac{8}{2^3}+a=-1+a=a-1.\)

The normal has equation

\(x+4y=b.\)

Rearranging gives

\(y=-\dfrac14x+\dfrac{b}{4},\)

so the gradient of the normal is \(-\dfrac14\).

The tangent and normal are perpendicular, so their gradients multiply to \(-1\). Therefore the tangent gradient must be \(4\).

Hence

\(a-1=4,\)

so

\(a=5.\)

Now find the point on the curve when \(x=2\):

\(y=\dfrac{4}{2^2}+5(2)+7=1+10+7=18.\)

Substitute \((2,18)\) into \(x+4y=b\):

\(b=2+4(18)=74.\)

Therefore

\(a=5,\quad b=74.\)