Starting from the condition for bright fringes we get

$$
frac{1}{2}lambda + 2d = mlambda
$$
yielding for $m$

$$
m=frac{frac{1}{2}lambda + 2d}{lambda}.
$$
If we put $d=t$ i.e. the distance between the plates equal to the thickness of the hair we get the condition for the last fringe and $m$ there is obviously equal to the total number of fringes.

$$
m=frac{frac{1}{2}lambda + 2t}{lambda}.
$$
We see that by increasing $t$ we increase the numerator which means that $m$ i.e. the maximum number of bright fringes increases. Now putting $t=80.0text{ $mu$m}$ we obtain

$$
m=frac{frac{1}{2}times741times10^{-9}text{ m} + 2times80.0times10^{-6}text{ m}}{741times10^{-9}text{ m}} approx 216.
$$

Starting from the condition for bright fringes we get

$$
frac{1}{2}lambda + 2d = mlambda
$$
yielding for $m$

$$
m=frac{frac{1}{2}lambda + 2d}{lambda}.
$$
If we put $d=t$ i.e. the distance between the plates equal to the thickness of the hair we get the condition for the last fringe and $m$ there is obviously equal to the total number of fringes.

$$
m=frac{frac{1}{2}lambda + 2t}{lambda} = frac{1}{2}+frac{2t}{lambda}.
$$
We see that by decreasing $lambda$ (which is effectively done when we pass from red to blue light) we decrease the denominator in the second term on the right which means that $m$ i.e. the maximum number of bright fringes increases.

We need to see which condition is satisfied, the one for constructive (meaning that there is a bright fringe) or the one for destructive interference (meaning that there is a dark fringe).
This is determined by the quantity

$$
frac{1}{2}lambda+2d = frac{1}{2}times 550times10^{-9}text{ m} + 2times7.7times10^{ -5}text{ m}=1.54278times10^{ -4}text{ m}.
$$

Wee see that this is exactly
$$
280.5lambda=left(280+frac{1}{2}right)lambda
$$
which is the condition for destructive interference i.e. there is a dark fringe.