Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have here two wavelengths, $\lambda$ and $\lambda+\Delta \lambda$.
At a small angle approximation, $\sin\theta\approx \tan \theta$, where the bright fringes in a diffraction grating are given by
$$d\sin\theta_m=m\lambda$$
and
$$y_m=L\tan\theta$$
Hence,
$$y_m=L\sin\theta=\dfrac{m\lambda L}{d}$$
which is the same formula for the bright fringes in a double-slit.
So the first-order bright fringe for both is given by
$$y_1=\dfrac{\lambda L}{d}\tag 1$$
$$y_1'=\dfrac{(\lambda +\Delta \lambda)L}{d}$$
Thus, the separation of the first-order peaks is then given by
$$\Delta y=y_1'-y_1=\dfrac{(\lambda +\Delta \lambda)L}{d}-\dfrac{\lambda L}{d}$$
$$\boxed{\Delta y= \dfrac{ \Delta \lambda L}{d} }$$
$$\color{blue}{\bf [b]}$$
From the given graph, we can see that the width is the distance from the point at which the intensity is half the maximum intensity to the point opposite to it.
In a double-slit experiment, we know that the intensity is given by
$$I=I_{max}\cos^2\left(\dfrac{\pi yd}{L\lambda}\right)$$
where $I_{max}=4I_1$
So the point at which the intensity is half the maximum, $I=2I_1$, and $y=y'$ where $y'$ is the distance at this point.
$$2I_1=4I_1\cos^2\left(\dfrac{\pi y'd}{L\lambda}\right)$$
So that
$$\cos^2\left(\dfrac{\pi y'd}{L\lambda}\right)=\dfrac{1}{2}$$
Taking the square root for both sides,
$$\cos\left(\dfrac{\pi y'd}{L\lambda}\right)=\dfrac{1}{\sqrt2}$$
$$\dfrac{\pi y'd}{L\lambda}=\cos^{-1}\left[\dfrac{1}{\sqrt2} \right]=\dfrac{\pi}{4}$$
Thus,
$$y'=\dfrac{L\lambda}{4d}$$
where $y'=\frac{w}{2}$
$$\frac{w}{2}=\dfrac{L\lambda}{4d}$$
$$w=\dfrac{L\lambda}{2d} $$
where we know that the location of $m=1$ in a double-slit is $y_1=\dfrac{\lambda L}{d}$, so
$$\boxed{w=\dfrac{y_1}{2}}$$
where $N=2$ in a double-slit experiment.
$$\color{blue}{\bf [c]}$$
From the final result in part (b), we can say that
$$w=\dfrac{y_1}{N}\tag 2$$
The diffraction fringes can barely be resolved means that
$$w=\Delta y_{\rm min}$$
where $\Delta y_{\rm min}$ is found in part (a) above, the boxed formula.
$$w=\dfrac{ \Delta \lambda_{\rm min} L}{d}$$
Plugging from (2);
$$\dfrac{y_1}{N}=\dfrac{ \Delta \lambda_{\rm min} L}{d}$$
So that
$$\Delta \lambda_{\rm min}=\dfrac{dy_1}{NL}$$
Plugging from (1);
$$\Delta \lambda_{\rm min}=\dfrac{\lambda L}{d}\dfrac{d }{NL}$$
Therefore,
$$\boxed{\Delta \lambda_{\rm min}=\dfrac{\lambda}{N}}$$
$$\color{blue}{\bf [d]}$$
Plugging the given into the previous boxed formula after solving for $N$,
$$N=\dfrac{\lambda}{\Delta \lambda_{\rm min}}$$
$$N=\dfrac{ 656.27\times 10^{-9}}{(656.45-656.27)\times 10^{-9}}=\color{red}{\bf 3646}\;\rm lines$$
