Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can estimate that the index of refraction of 656 nm in the flint glass is
$$n_{656\rm \;nm}=\color{red}{\bf1.566}$$
and the index of refraction of 486 nm in the flint glass is
$$n_{486\rm \;nm}=\color{red}{\bf 1.587}$$
$$\color{blue}{\bf [b]}$$
According to Snell's law,
$$n_1\sin\theta_1=n_2\sin\theta_2$$
where all the angles are measured relative to the normal.
For the red light in flint glass,
$$n_{\rm flint glass,red}\sin\theta_{\rm flint glass}=n_{air}\sin\theta_{\rm air,red}$$
So,
$$ \theta_{\rm air,red}=\sin^{-1}\left[\dfrac{n_{\rm flint glass,red}\sin\theta_{\rm flint glass}}{n_{air}}\right]$$
Plugging the known;
$$ \theta_{\rm air,red}=\sin^{-1}\left[\dfrac{(1.566)\sin35^\circ}{1.0}\right]=\bf 63.9^\circ$$
For the blue light in flint glass,
$$ \theta_{\rm air,blue}=\sin^{-1}\left[\dfrac{n_{\rm flint glass,blue}\sin\theta_{\rm flint glass}}{n_{air}}\right]$$
Plugging the known;
$$ \theta_{\rm air,blue}=\sin^{-1}\left[\dfrac{(1.587)\sin35^\circ}{1.0}\right]=\bf 65.5^\circ$$
Therefore, the angle between the red and blue light, as it leaves the prism, is given by
$$\Delta \theta=\theta_{\rm air,blue}-\theta_{\rm air,red}$$
$$\Delta \theta=65.5^\circ-63.9^\circ=\color{red}{\bf 1.6^\circ}$$
Noting that your answer may differ from that since the graph is not perfectly accurate.
