Can someone explain this task to me?

a) The function with which the covered area can be calculated as a function of time is given.

The covered area is asked at two specific times. So you only have to use these times and calculate the areas:

b(0) = -0,06*0 + 0,6*0 + 0,8*0 +2 = 2
Answer: At the beginning of observation, the covered area is 2 cm^2

Now the second time. We have to convert the data into a number of commas in hours and minutes:
2 h 40 min = 2 h + 40/60 h = 2.67 h

b(2,67) = -0,06*2,67^3 + 0,6*2,67^2 + 0,8*2,67 +2 = 7.3 cm^2

(b) You need the graph first. Where you got it, I don’t know. Either there is an image for the task, or you can use a function plotter or you have to draw the graph yourself via a value table. He looks like this.

From the graph we read: the largest area is at t = 7.25 h = 7 h 15 min and the covered area is then 16.5 cm^2

(c) The bacterial culture has died when the covered area is 0.

It is then valid:
-0.06*t^3 + 0.6*t^2 + 0.8*t +2 = 0

I don’t know how to calculate this, but the graph shows
t = 11,4

(d) Then I look back in the graphs when the curve exceeds the 15 and when it falls below again:

t1 = 5.7 h = 5 h 0.7 * 60 min = 5 h 42 min
t2 = 8.7 h = 8 h 42 min

Solution The bacterial culture covers an area of more than 15 cm2 between 5 h 42 min and 8 h 42 min.