Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, in a diffraction grating, that the bright fringe angle is given by
$$d\sin\theta=m\lambda$$
So,
$$\lambda=\dfrac{d\sin\theta}{m}\tag 1$$
For a small change in wavelength of $\Delta \lambda$, the diffraction angle changes by $\Delta \theta$.
We can assume that $\dfrac{\Delta \lambda}{\Delta\theta}=\dfrac{d\lambda}{d\theta}$ when the changes are tiny.
Differentiating (2) to $d/d\theta$;
$$\dfrac{d\lambda}{d\theta}=\dfrac{d}{d\theta}\dfrac{d\sin\theta}{m}=\dfrac{d\cos\theta}{m}$$
Hence,
$$\dfrac{\Delta \lambda}{\Delta\theta}=\dfrac{d\cos\theta}{m}$$
where $\cos\theta=\sqrt{1-\sin^2\theta}$ [recall that $\sin^\theta+\cos^2\theta=1$].
$$\dfrac{\Delta \lambda}{\Delta\theta}=\dfrac{d\sqrt{1-\sin^2\theta}}{m}$$
Solving for $\Delta \theta$;
$$\Delta\theta=\dfrac{m\Delta \lambda}{d\sqrt{1-\sin^2\theta}}$$
where $\sin\theta=m\lambda/d$
$$\Delta\theta=\dfrac{ \Delta \lambda}{\dfrac{d}{m}\sqrt{1- \left(\dfrac{m\lambda}{d}\right)^2}}$$
$$\boxed{\Delta\theta=\dfrac{ \Delta \lambda}{\sqrt{\left(\dfrac{d}{m}\right)^2- \lambda^2}}}$$
$$\color{blue}{\bf [b]}$$
We just need to plug the given into the boxed formula above, where $d=1/N$,
$$ \Delta\theta=\dfrac{ \Delta \lambda}{\sqrt{\left(\dfrac{1}{Nm}\right)^2- \lambda^2}}$$
For the first-order angular separation where $m=1$,
$$ \Delta\theta_1=\dfrac{ (589.6-589)\times 10^{-9}}{\sqrt{\left(\dfrac{10^{-3}}{(600)(1)}\right)^2- (589\times 10^{-9})^2}}$$
$$\Delta \theta_1=\bf 3.85\times 10^{-4}\;\rm rad=\color{red}{\bf 0.022}^\circ$$
For the second-order angular separation where $m=2$,
$$ \Delta\theta_2=\dfrac{ (589.6-589)\times 10^{-9}}{\sqrt{\left(\dfrac{10^{-3}}{(600)(2)}\right)^2- (589\times 10^{-9})^2}}$$
$$\Delta \theta_2=\bf 1.02\times 10^{-2}\;\rm rad=\color{red}{\bf 0.058}^\circ$$
