Answer: \(\alpha=5\), \(\beta=6\); unit vector \(\dfrac{-8\mathbf i+2\mathbf j}{\sqrt{68}}\).

(a) First find \(4\mathbf a-\mathbf b\):

\(4\mathbf a-\mathbf b =4(\alpha\mathbf i+\mathbf j)-(12\mathbf i+\beta\mathbf j).\)

So

\(4\mathbf a-\mathbf b=(4\alpha-12)\mathbf i+(4-\beta)\mathbf j.\)

This is equal to

\((\alpha+3)\mathbf i-2\mathbf j.\)

Equating the \(\mathbf i\)-components,

\(4\alpha-12=\alpha+3.\)

Hence

\(3\alpha=15,\)

so

\(\alpha=5.\)

Equating the \(\mathbf j\)-components,

\(4-\beta=-2.\)

Hence

\(\beta=6.\)

(b) Since

\(4\mathbf a-\mathbf b=8\mathbf i-2\mathbf j,\)

we have

\(\mathbf b-4\mathbf a=-8\mathbf i+2\mathbf j.\)

The magnitude is

\(\sqrt{(-8)^2+2^2}=\sqrt{68}.\)

Therefore the unit vector in this direction is

\(\frac{-8\mathbf i+2\mathbf j}{\sqrt{68}}.\)