Looks like [tex]n_t[/tex] is the number of subintervals you have to use with the trapezoidal rule, and [tex]n_s[/tex] for Simpson's rule. In the attachments, I take both numbers to be 4 to make drawing simpler.
Split up the integration interval [1, 8] into n subintervals. Each subinterval then has length (8 - 1)/n = 7/n. This gives us the partition
[1, 1 + 7/n], [1 + 7/n, 1 + 14/n], [1 + 14/n, 1 + 21/n], ..., [1 + 7(n - 1)/n), 8]
The left endpoint of the [tex]i[/tex]th interval is given by the arithmetic sequence,
[tex]\ell_i=1+\dfrac{7(i-1)}n[/tex]
and the right endpoint is
[tex]r_i=1+\dfrac{7i}n[/tex]
both with [tex]1\le i\le n[/tex].
For Simpson's rule, we'll also need to find the midpoints of each subinterval; these are
[tex]m_i=\dfrac{\ell_i+r_i}2=1+\dfrac{7(2i-1)}{2n}[/tex]
The area under the curve is approximated by the area of 12 trapezoids. The partition is (roughly)
[1, 1.58], [1.58, 2.17], [2.17, 2.75], [2.75, 3.33], ..., [7.42, 8]
The area [tex]A_i[/tex] of the [tex]i[/tex]th trapezoid is equal to
[tex]A_i=\dfrac{f(r_i)+f(\ell_i)}2(r_i-\ell_i)[/tex]
Then the area under the curve is approximately
[tex]\displaystyle\int_1^8f(x)\,\mathrm dx\approx\sum_{i=1}^{12}A_i=\frac7{24}\sum_{i=1}^{12}f(\ell_i)+f(r_i)[/tex]
You first need to use the graph to estimate each value of [tex]f(\ell_i)[/tex] and [tex]f(r_i)[/tex].
For example, [tex]f(1)\approx2.1[/tex] and [tex]f(1.58)\approx2.2[/tex]. So the first subinterval contributes an area of
[tex]A_1=\dfrac{f(1.58)+f(1)}2(1.58-1)=1.25417[/tex]
For all 12 subintervals, you should get an approximate total area of about 15.9542.
Over each subinterval, we interpolate [tex]f(x)[/tex] by a quadratic polynomial that passes through the corresponding endpoints [tex]\ell_i[/tex] and [tex]r_i[/tex] as well as the midpoint [tex]m_i[/tex]. With [tex]n=24[/tex], we use the (rough) partition
[1, 1.29], [1.29, 1.58], [1.58, 1.88], [1.88, 2.17], ..., [7.71, 8]
On the [tex]i[/tex]th subinterval, we approximate [tex]f(x)[/tex] by
[tex]L_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}[/tex]
(This is known as the Lagrange interpolation formula.)
Then the area over the [tex]i[/tex]th subinterval is approximately
[tex]\displaystyle\int_{\ell_i}^{r_i}f(x)\,\mathrm dx\approx\int_{\ell_i}^{r_i}L_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6\left(f(\ell_i)+4f(m_i)+f(r_i)\right)[/tex]
As an example, on the first subinterval we have [tex]f(1)\approx2.1[/tex] and [tex]f(1.29)\approx1.9[/tex]. The midpoint is roughly [tex]m_1=1.15[/tex], and [tex]f(1.15)\approx2[/tex]. Then
[tex]\displaystyle\int_{\ell_1}^{r_1}f(x)\,\mathrm dx\approx\frac{1.29-1}6(2.1+4\cdot2+1.9)=0.58[/tex]
Do the same thing for each subinterval, then get the total. I don't have the inclination to figure out the 60+ sampling points' values, so I'll leave that to you. (24 subintervals is a bit excessive)
For part 2, the average rate of change of [tex]f(x)[/tex] between the points D and F is roughly
[tex]\dfrac{f(5.1)-f(2.7)}{5.1-2.7}\approx\dfrac{1.3-2.6}{5.1-2.7}\approx-0.54[/tex]
where 5.1 and 2.7 are the x-coordinates of the points F and D, respectively. I'm not entirely sure what the rest of the question is asking for, however...
