Calculus help!?

8/n  

seems correct 

but 

remember the interval start at x =1  

so 1st interval goes from 1  to 1 + 8/n     

so right endpoint is  1  + 8/n  

then you second answer(b)  should 

1 + (8/n), 1 + (16/n) , 1+ (24/n)  

3rd answer (c) 

1 + (8k/n)  

4th answer  (d)

(1 + (8k/n))^2  

You could multiply it out 

(1  +  2(8k/n)   + 64k^2/n^2   ) 

answer multiplied out 

64k^2/n^2  + 16k/n   +  1 

5th answer  (e)

(64k^2/n^2 + 16k/n + 1 ) *8/n  

(e)

256k^2/n   + 128k  + 8/n   

That hint ("Is 8/n in the interval...?") doesn't show in the image, and I see a "part b"...but no part 2.

The value 8/n is the ∆x value you found in part a.  It's the distance between consecutive x_k values, not the x coordinate of any value.  It's the size of a step that will take you from the left endpoint at x=1 to the right at x=9 in n equal steps.

The first subinterval begins at x=1 on the left, and ends at x=1+∆x on the right.  So it's [1, 1+∆x].  The second subinterval begins where the first ended (x = 1+∆x) and ends at x=1 + 2∆x.  That's two steps from the left end of the full interval.  Continue that pattern and you should see that the k'th interval (for k = 1, 2, 3, ..., n) begins at 1 + (k-1)*∆x and ends at 1 + k*∆x.

In general, on the interval [a, b] with n equal subintervals, you have a constant ∆x value of (b-a)/n; and the k'th interval is [a + (k-1)∆x, a + k*∆x]

I suggest you derive that for yourself, on paper, and then check with the derivation in your notes or textbook.  Dealing with subintervals, and knowing the difference between a displacement and a position, are fundamental ideas. You need to get used to them, or be overwhelmed by the repeated use you're going to need to make of them in the next couple of semesters.

Then get back to how that works with [1,9] as a full interval with, say, k=4 subintervals.  Once it "clicks" for you, you'll be surprised at how hard this isn't.