Answer
$782\;\rm slit/cm$
$658\;\rm nm$
Work Step by Step
We know, for bright fringes in a diffraction grating, that
$$m\lambda=d\sin\theta$$
For the first peak above the maxima $m=1$, so
$$ \lambda=d\sin\theta\tag 1$$
And we know that the distance from the center fringe to the first-order bright fringe is given by
$$x=L\tan\theta $$
Thus, the angle is given by
$$\theta=\tan^{-1}\left[ \dfrac{x}{L}\right]$$
Therefore, the angle of the first peak above the center maxima of the first wavelength is
$$\theta_1=\tan^{-1}\left[ \dfrac{x_1}{L}\right]=\tan^{-1}\left[\dfrac{3.32}{72}\right]=\bf 2.640^\circ$$
And the angle of the first peak above the center maxima of the second wavelength is
$$\theta_2=\tan^{-1}\left[ \dfrac{x_2}{L}\right]=\tan^{-1}\left[\dfrac{3.71}{72}\right]=\bf 2.95^\circ$$
From (1), for $\lambda_1$
$$d=\dfrac{\lambda_1}{\sin\theta_1}$$
And hence, the number of slits per centimeter is given by
$$N=\dfrac{1}{d}=\dfrac{\sin\theta_1}{\lambda_1}$$
Plugging the known;
$$N=\dfrac{\sin2.640^\circ}{589\times10^{-9}}=78201\;\rm slit/m$$
$$N=\color{red}{\bf 782}\;\rm slit/cm$$
Solving (1) to find the second wavelength,
$$\lambda_2=d\sin\theta_2=\dfrac{1}{78201}\times \sin 2.95^\circ=\color{red}{\bf 658\times10^{-9}}\;\rm m $$
