Answer
(a) longer than
(b) $0.63\mu m$
Work Step by Step
(a) We know that $\lambda\frac{1}{m}$. This shows that if $m$ is increased from $10$ to $7$, then the wavelength will be increased and hence the new wavelength will be greater than $440nm$.
(b) We have $\theta=tan^{-1}(\frac{y}{L})$
$\theta=\tan^{-1}(\frac{0.12m}{2.3m})=3.0^{\circ}$
Now $\lambda=\frac{d}{m}sin\theta$
We plug in the known values to obtain:
$\lambda=\frac{8.5\times 10^{-5}m}{7}sin 3.0^{\circ}=0.63\mu m$