A projectile is fired directly upward with a muzzle velocity of 860 ft per second from a height of 7 ft above?

without data about drag, I presume the idea is that this is done in a vacuum on earth.
a. a = 32 ft/sec^2
     v = 860 ft/sec - (32 ft/sec^2 * t sec) or (860 - 32t)(ft/sec^2)
b. s = 7 ft + integral from 0 to t of v dt
    integral = 860t (ft/sec) - 16t^2 (ft/sec^2)
     100 = 7 + 860t - 16 t^2    -> 16 t^2 - 860t + 93 = 0
   discard the larger solution, which is the time till it comes back down to 100 ft.  
   t = 0.10836 seconds
c. calculate time going up, add time going down,
     which is longer, because it travels extra seven feet
    on the way down.
    stops going up when v = 0.
    860 = 32t
    t = 26.875 seconds
    h = 7 + 860t - 16t^2 = 11563.25 feet
    t(up) = 26.875 seconds.
    t(down) = 26.88314 seconds
    t(total) = 53.75814 seconds

When you do projectile questions there is only one thing you need to remember and that is gravity

Downward acceleration is by gravity = 9.8ms?² = 32ft s?²

This movement is only up and down so we only need to look at the vertical.

a(t) = -32 .... (negative as it goes down)

now to find velocity we integrate with respect to time
v(t) = ?-32 dt
= C - 32t

Find the constant of integration using the initial comditions - muzzle velocity = 860ft

v(0) = C - 32(0)
860 = C
v(t) = 860 - 32t

integrate again to find a displament formula

s(t) = 860t - 16t² + C

again find the constant of intgration using the initial conditions.  Now even though the projectile starts 7 feet above the ground it is easier if we call the firing point (0, 0) so s(0) = 0

s(0) = 860(0) - 16(0)² + C
0 = C
s(t) = 860t - 16t²

NOTE: If we had the ground below as the zero point then C would be 7 which introduces an extra term in the formula which is why I make the firing point (0, 0)

......
In summary

a(t) = -32
v(t) = 860 - 32t
s(t) = 860t - 16t²
......
------------------------------
Using these we can solve all of your problems

a. Determine a function for the height of the projectile t seconds after its released.

s(t) = 860t - 16t²

------------------------------

b. how long does it take the projectile to reach a height of 100 ft on its way up?

so in other words "find first t when s(t) = 93ftstance of firing point is 7 feet above the ground)

93 = 860t - 16t²
16t² - 860t + 93 = 0
t = (860 ± ?(860² - 4(93)(16))) / (2*16)
= (860 ± 4?(215² - (93)(4))) / (2*16)
= (215 ± ?45853) / 8

The first time it hits 100ft above the ground
t = (215 - ?45853) / 8 ... (The "+" on the way down)
0.10836 sec (5dp)

.....................
Aside:
If they actualy meant 100ft above the firing point (which I highly doubt):
s(t) = 860t - 16t²
100 = 860t - 16t²
25 = 215t - 4t²
4t² - 215t + 25 = 0
t = (215 ± ?(215² - 4(25)(4))) / (2*16)
t = (215 ± ?45825) / 32
The first time it hits 100ft above the firing point
t = (215 - ?45825) / 32
t = 0.029133 sec (6dp)
------------------------------

c. How long is the projectile in the air?
It lands when s(t) = -7 ... (7 ft below firing point(
s(t) = 860t - 16t²
-7 = 860t - 16t²
16t² - 860t - 7 = 0
t = (860 ± ?(860² - 4(-7)(16))) / (2*16)
= (860 ± 4?(215² + (7)(4))) / (2*16)
= (215 ± ?46253) / 8
It's the "+" because the "-" will be before take off dur to the parabolic shape of projectiles
t = (215 + ?46253) / 8
= (215 + ?46253) / 8
= 53.758sec (3dp)