For the Trapezoidal Rule, you actually use $n+1$ points. For example, in the simple case where you are integrating $f(x)$ from $0$ to $1$, and you want $T_4$, you evaluate $f$ at the points $0/4$, $1/4$, $2/4$, $3/4$, and $4/4$. It is $n+1$ points because we use the endpoints.
For the Midpoint Rule, you use $n$ points, but these are not the same points as for the Trapezoidal Rule. They are the midpoints of our intervals. So in the example discussed above, for $M_4$ you would be evaluating $f$ at $1/8$, $3/8$, $5/8$, and $7/8$.
The Simpson Rule $S_{2n}$ uses evaluation of $f$ at $2n+1$ points. If for example $n=4$, then you are dividing the interval into $8$ subintervals. With $n=4$ and the interval $[0,1]$, you would be using the points $0/8$, $1/8$, $2/8$, $3/8$, $4/8$, $5/8$, $6/8$, $7/8$, and $8/8$.
Note that $1/8$, $3/8$, $5/8$ and $7/8$ are the points that were used for $M_4$. The points $0/8$, $2/8$, $4/8$, $6/8$, and $8/8$ are just $0/4$, $1/4$, $2/4$, $3/4$, and $4/4$, exactly the points that were used for $T_4$.
A more abstract summary: $T_n$ uses $n+1$ points, and $M_n$ uses $n$ points. But the $n$ points used by $M_n$ are completely different from the points used for $T_n$. So altogether, $T_n$ and $M_n$ carry information about function evaluation at $2n+1$ points, which is exactly what $S_{2n}$ does.
I have not written out a proof of the formula, only tried to deal with your discomfort with the $2n$ on one side and $n$'s on the other. The formula is not hard to verify. Let's do it explicitly for $n=4$. Write down, say for the interval $[0,1]$, what $T_4$ is.
We have
$$T_4=\frac{1}{8}(f(0)+2f(1/4)+2f(1/2)+2f(3/4)+f(1)).$$
Now write down $M_4$:
$$M_4=\frac{1}{4}(f(1/8)+f(3/8)+f(5/8)+f(7/8)).$$
Now calculate $T_4+2M_4$. It is convenient for the addition to make sure that $M_4$ has denominator $8$, so write $2M_4$ as $\frac{1}{8}(4f(1/8)+4f(3/8)+4f(5/8)+4f(7/8))$, and add. Divide by $3$ and you will get the expression you would get in $S_{8}$. The same method works in general.
The average of the left and right endpoint estimates, sometimes called $L_n$ and $R_n$ respectively, is actually the same as the trapezoid estimate, called sometimes $T_n$, which is based on drawing trapezoidal slabs based on the values of the function at the two endpoints.
One would not expect the midpoint rule $M_n$ to agree in general with any of the above, since it uses sample values of the function at the midpoints of the subintervals, and those values are independent (in general) of the values of the function at the endpoints of the subintervals.
ADDED: I see now what they did in the "midpoint rule" calculation. For it they actually used three intervals of length $2$, so that the values at the midpoints are actually on the table of values given, and are (from left to right) the numbers $8,38,56$ with a sum of $102$, which must be multiplied by the subinterval length of $2$ to get the answer of $204$ as stated in the printout. (By the way I had to zoom in a bit to see that.)
Best Answer
What are the given correct answers?
Your work seems correct. But of course $n$ is a positive integer, so you need to round your values up to the ceiling of your calculated answers. So $n=248$ guarantees the desired precision for the trapezoidal rule and $n=175$ does so for the midpoint rule.
However, those values were gotten by using the formulas that guarantee the desired precision, given a bound on the absolute value of the second derivative. You may get much better precision than the guaranteed value. Using $n=175$ I get the error $0.000028237574$ using the midpoint rule, much better than the desired $0.001$. You can get the desired error with much smaller values of $n$.
You should repeat the calculations of the approximate integrals to find just which value of $n$ actually gives you the desired error. For the midpoint rule in your situation, I get $n=93$ as the smallest $n$, which give the error $0.000099982028$. ($n=92$ gives $0.000102167251$.) For the trapezoidal rule I get $n=132$, giving the error $0.000099262595$.
Is the given correct answer for the midpoint rule $n=175$, my theoretical answer, or $n=93$, my calculated answer, or something else?