Answer
$462\;\rm nm$, $676\;\rm nm$
Work Step by Step
For the first spectrum line, we assume that the diffraction angle is half the difference between the angles on opposite sides of the center
Thus, the angle is given by
$$\theta_1=\dfrac{\theta_{up}-\theta_{below}}{2}\tag 1$$
And we know, for diffraction gratings, that
$$\overbrace{d}^{\frac{1}{N}}\sin\theta_1=\overbrace{m}^{=1}\;\lambda_1$$
Hence,
$$\lambda_1=d\sin\theta_1=\dfrac{\sin \theta_1}{N}$$
Plugging from (1);
$$\lambda_1= \dfrac{\sin \left[\dfrac{\theta_{up}-\theta_{below}}{2}\right]}{N}$$
Plugging the known;
$$\lambda_1= \dfrac{\sin \left[\dfrac{26^\circ 38'-(-26^\circ 18')}{2}\right]\times 10^{-2}}{ 9650}=461.8\times 10^{-9}\;\rm m$$
$$\lambda_1\approx \color{red}{\bf 462}\;\rm nm$$
---
For the other spectrum line, we also assume that the diffraction angle is half the difference between the angles on opposite sides of the center
Thus, the angle is given by
$$\theta_2=\dfrac{\theta_{up}-\theta_{below}}{2}\tag 2$$
And we know, for diffraction gratings, that
$$\overbrace{d}^{\frac{1}{N}}\sin\theta=\overbrace{m}^{=1}\;\lambda_2$$
Hence,
$$\lambda_2=d\sin\theta_2=\dfrac{\sin \theta_2}{N}$$
Plugging from (2);
$$\lambda_2= \dfrac{\sin \left[\dfrac{\theta_{up}-\theta_{below}}{2}\right]}{N}$$
Plugging the known;
$$\lambda_2= \dfrac{\sin \left[\dfrac{41^\circ 02'-(-40^\circ 27')}{2}\right]\times 10^{-2}}{ 9650}=676.3\times 10^{-9}\;\rm m$$
$$\lambda_2\approx \color{red}{\bf 676}\;\rm nm$$
