The "appropriate" sample points are: $1+1/n, 1+2/n,1+3/n,\ldots,2$ for the sum on the left and $1 , 1+1/n,1+2/n,\ldots,1-(n-1)/n$ for the sum on the right.
The function $f(t)={1\over t}$ is decreasing over $[1,2]$.
To obtain a lower estimate of the integral, we take the sum of the areas of the rectangles $A_1, A_2, \ldots, A_n,$ where $A_i$ is the rectangle whose base is $[1+(i-1)/n, 1+i/n]$ and whose height is $f(1+i/n)={1\over 1+{i\over n}}$ ($f$ evaluated at the right endpoint of $A_i$). This gives the underestimate $ L=\sum_{i=1}^n\underbrace{ {1\over 1+{i\over n}}}_{{\rm height\ of\ }A_i }\cdot \underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i}. $
So $L$ is the sum on the left in your post.
By taking the height of $A_i$ to be $f(1+(i-1)i/n)={1\over 1+{(i-1)\over n}}$ ($f$ evaluated at the left endpoint of $A_i$), we obtain the over estimate: $U=\sum_{i=1}^n\underbrace{ {1\over 1+{i-1\over n}}}_{{\rm height\ of\ }A_i }\cdot \underbrace{1\over n}_{{\rm width\ of\ }A_i }=\sum_{i=1}^n {1\over n+i-1} =\sum_{i=0}^{n-1} {1\over n+i } . $
So $U$ is the sum on the right in your post.
For part $b)$, using the above, if you estimate the integral with one of these sums, the error $E$ will satisfy: $ |E|\le U-L=\sum_{i=0}^{n-1} {1\over n+i }-\sum_{i=1}^n {1\over n+i}={1\over n}-{1\over 2n}={1\over 2n} $ So you just need to find the smallest value of $n$ that satisfies ${1\over 2n}\le5\cdot10^{-6}$.