Answer
a) $1.26\;\rm m$
b) $7\;\rm bright\;fringes$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the distance between the two $m=1$ bright fringes is twice the distance of $y_1$.
$$\Delta y=2y_1\tag 1$$
where $y_1$ is the distance of the $m=1$ bright fringe.
Recalling that $y_1$ in a diffraction grating is given by
$$y_1=L\tan\theta_1\tag 2$$
Now we need to find the angle which is given by
$$d\sin\theta_1=(1)\lambda$$
Hence,
$$\theta_1=\sin^{-1}\left( \dfrac{\lambda} {d}\right)$$
where $d=1/N$,
$$\theta_1=\sin^{-1}\left( {N \lambda} \right)$$
Plugging into (2);
$$y_1=L\tan\left[ \sin^{-1}\left( {N \lambda} \right)\right]$$
Plugging into (1);
$$\Delta y=2L\tan\left[ \sin^{-1}\left( {N \lambda} \right)\right]$$
Plugging the known;
$$\Delta y=2(2)\tan\left[ \sin^{-1}\left( {\frac{600}{10^{-3}} (500\times 10^{-9})} \right)\right]$$
$$\Delta y=\color{red}{\bf 1.26}\;\rm m$$
$$\color{blue}{\bf [b]}$$
We know that the number of bright fringes is determined by the maximum value of $m$,
$$\sin\theta_m=\dfrac{m\lambda}{d}= m\lambda N $$
where the right side must be less than 1, so
$$ \dfrac{m\lambda}{d}= m(500\times 10^{-9})(600\times 10^3)$$
$$ \dfrac{m\lambda}{d}=\dfrac{3}{10} m $$
At $m=1$,
$$ \dfrac{m\lambda}{d}=\dfrac{3}{10} (1)=\bf 0.3\lt 1 $$
At $m=2$,
$$ \dfrac{m\lambda}{d}=\dfrac{3}{10} (2)=\bf 0.6\lt 1 $$
At $m=3$,
$$ \dfrac{m\lambda}{d}=\dfrac{3}{10} (3)=\bf 0.9\lt 1 $$
At $m=4$,
$$ \dfrac{m\lambda}{d}=\dfrac{3}{10} (4)= 1.2\gt 1 $$
So we will get a maximum number of $m=3$.
Hence, the total number of bright fringes is 3 on each side (which means 6 bright fringes) plus the central maximum.
Therefore, we can get 7 bright fringes only.
