Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First of all, we need to sketch this problem, as we see below.
We can see that there are two path length differences now due to the tilting of the laser beam.
$$\Delta r_1=d\sin\theta_m$$
and
$$\Delta r_2=d\sin\phi$$
Then the total path length difference between the two rays is then given by an integer number of light wavelengths.
$$\Delta r_1+\Delta r_2=m\lambda$$
Plugging from the two previous formulas,
$$d\sin\theta_m+d\sin\phi=m\lambda$$
Thus,
$$\boxed{d(\sin\theta_m+\sin\phi)=m\lambda}\tag{$m=0,\pm1,\pm2.\pm3,..$}$$
$$\color{blue}{\bf [a]}$$
Solving the previous formula for $\theta_m$,
$$\theta_m=\sin^{-1}\left[\dfrac{m\lambda}{d}-\sin\phi\right]$$
where $d=1/N$,
$$\theta_m=\sin^{-1}\left[ m N\lambda -\sin\phi\right]$$
At $m=1$,
$$\theta_{(1)}=\sin^{-1}\left[ (1)\left(\frac{600}{10^{-3}}\right) (500\times 10^{-9}) -\sin30^\circ\right]$$
$$\theta_{(1)}=\color{red}{\bf -11.5}^\circ$$
At $m=2$,
$$\theta_{(2)}=\sin^{-1}\left[ (2)\left(\frac{600}{10^{-3}}\right) (500\times 10^{-9}) -\sin30^\circ\right]$$
$$\theta_{(2)}=\color{red}{\bf -53.1}^\circ$$
