Assuming #(fg)(7)# is shorthand for #f(g(7))#, what this asks you to do is to first evaluate the function #g(x)# for #x = 7#, and then shove the result of that function into #f(x)#, to get the final result.
When not in a rush, I personally like to do this generally first (which means purely using variables), and then solve for the specific value later, just in case it will be needed to be solved for other values too.
Here's what we have:
#f(x) = x^2 - 5x#
#g(x) = 8 - x#
And let's say #h(x) = f(g(x))#.
Let's start by replacing #g(x)# with its value:
#h(x) = f(g(x)) = f(8 - x)#
Now, the #x# in #f(x)# is actually whatever it takes in. In this case, the input is #g(x)#, or #8 - x#. Continuing:
#h(x) = f(8 - x) = (8 - x)^2 - 5(8 - x)#
Here, what we did was replace every instance of #x# in #f(x)# with the actual input, #(8 - x)#.
Notice, #f(x)# was defined as #x^2 - 5x#, but here, because the input is #(8 - x)#, we have #(8 - x)^2 - 5(8 - x)# instead.
Time to simplify:
#h(x) = (8 - x)^2 - 5(8 - x) = (8 - x)^2 - 40 + 5x#
OK, so we expanded #- 5(8 - x)# first... now we'll expand #(8 - x)^2#:
#= (8 - x)(8 - x) - 40 + 5x#
#= 64 - 8x - 8x + x^2 - 40 + 5x#
Combine like terms:
#= 64 - 16x + x^2 - 40 + 5x#
#= 64 - 11x + x^2 - 40#
#= 24 - 11x + x^2#
Oops, let's rearrange the terms to make it look neater:
#= x^2 - 11x + 24#
Nice! #h(x) = x^2 - 11x + 24#. Now we're just asked to evaluate it at #x = 7#:
#h(7) = (7)^2 - 11(7) + 24#
#= 49 - 11(7) + 24#
#= 49 - 77 + 24#
#= 73 - 77#
#= -4#
