Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have here two cases,
$\bullet$ when the index of refraction is $n$, applying Snell's law
$$n_1\sin\theta_1=n_2\sin \theta_2 $$
where $n_1=1$ for air, and $n_2=n$,
$$ \sin\theta_1=n \sin \theta_2 \tag 1$$
$\bullet$ when the index of refraction is $n+\delta n$, applying Snell's law
$$n_1\sin\theta_1=n_2\sin(\theta_2 +\delta\theta)$$
where $n_1=1$ for air, and $n_2=n+\delta n$,
$$ \sin\theta_1=(n +\delta n)\sin(\theta_2+\delta\theta )\tag 2$$
It is obvious that the left sides of (1) and (2) are identical, and so do the right sides.
$$ n \sin \theta_2 =(n +\delta n)\sin(\theta_2+\delta\theta )$$
$$ n \sin \theta_2 =(n +\delta n)(\sin\theta_2 \cos\delta\theta+\cos\theta_2 \sin\delta \theta)$$
and since $\delta \theta\lt \lt \theta$, then $\cos\delta\theta\approx 1$, and $\sin\delta \theta\approx \delta \theta$ [Recall small angle approximation]
$$ n \sin \theta_2 =(n +\delta n)(\sin\theta_2 +\delta \theta\cos\theta_2 )$$
$$ n \sin \theta_2 = n\sin\theta_2 +n\delta \theta\cos\theta_2 +\delta n\sin\theta_2 +\delta n\delta \theta\cos\theta_2$$
$$\overbrace{ n \sin \theta_2 - n\sin\theta_2}^{=0} =n\delta \theta\cos\theta_2 +\delta n\sin\theta_2 +\delta n\delta \theta\cos\theta_2$$
$$0=n\delta \theta\cos\theta_2 +\delta n\sin\theta_2 +\overbrace{\delta n\delta \theta\cos\theta_2}^{\approx 0}$$
the last term approaches zero since it is a product of two small values.
Solving for $\delta \theta$,
$$ \delta \theta =-\dfrac{\delta n\sin\theta_2}{n\cos\theta_2}=-\dfrac{\delta n }{n}\tan\theta_2$$
$$ \boxed{\delta \theta =-\left[\dfrac{\delta n }{n}\right]\tan\theta_2}$$
$$\color{blue}{\bf [b]}$$
We need to use Snell's law in both cases,
For red,
$$n_{air}\sin30^\circ=n_{red}\sin\theta_{red}$$
$$ \sin30^\circ=1.552\sin\theta_{red}=\bf 18.794^\circ$$
Now we can use the boxed formula above, where $\delta\theta=-0.28^\circ\cdot\dfrac{\pi}{180}\;\rm rad$, $n=n_{red}=1.552$
$$\delta \theta =-\left[\dfrac{\delta n }{n}\right]\tan\theta_2$$
Solving for $\delta n$,
$$ \delta n=\dfrac{-n_{red}\delta \theta}{\tan\theta_{2,red}} $$
Plugging the known;
$$ \delta n=\dfrac{-(1.552)\dfrac{-0.28^\circ\pi}{180} }{\tan 18.794^\circ} =\bf 0.0223$$
Therefore,
$$n_{violet}=n_{red}+\delta n=1.552+0.0223$$
$$n_{violet}=\color{violet}{\bf 1.5743}$$
