Icannot solve this. i don't know how.

The notation f(x) means you have a function that has been given the name f, and it makes use of the variable x. The variable in the parentheses is called the "argument" of the function f.

(a) To find f(q), you put q everywhere x is in the function equation. This is called evaluating the function for an argument of "q". In the following, note that we have simply changed x to q. (It's really that simple.)

... f(q) = q² -2q +3

(b) As in the previous case, we replace x with (x+h) everywhere.

... f(x+h) = (x+h)² -2(x+h) +3

You can multiply it out, but there appears to be no need to do so for this part of the question.

(c) The intent here is that f(x+h) and f(x) will be replaced by their values and the whole thing simplified. This requires you  expand the expression you see in part (b), subtract f(x), collect terms, and divide the whole thing by h. You have to make use of what you know about multiplying binomials.

We can do it in parts:

... f(x+h) = (x+h)² -2(x+h) +3

... = (x² +2xh +h²) + (-2x -2h) +3

Separating the h terms, this looks like ...

... = (x² -2x +3) + (2xh -2h +h²)

Now, we can finish the numerator part of the expression by subtracting f(x):

... f(x+h) -f(x) = (x² -2x +3) +(2xh -2h +h²) -(x² -2x +3)

You can see that the stuff in the first parentheses matches that in the last parentheses, so when we subtract the latter from the former, we get zero. We are left with only the terms containing h.

... f(x+h) -f(x) = 2xh -2h +h²

To finish up this problem, we need to divide this numerator value by the denominator h.

... (f(x+h) -f(x))/h = (2xh -2h +h²)/h

... = (2xh)/h -(2h)/h +h²/h

... = 2x -2 +h . . . . . this is the value of the expression

... (f(x+h) -f(x))/h = 2x -2 +h