Answer
$$ M_6=0.88,\ \ \ \ T_6=0.883 $$
Work Step by Step
Given $$\int_{0}^{\pi/4} \sec xd x,\ \ \ N=6 $$
Since $\Delta x=\dfrac{b-a}{N}=\dfrac{\pi/4}{6}=\dfrac{\pi}{24} $
Therefore, the sub intervals consist of
$$[0,\pi/24],[\pi/24,2\pi/24],[2\pi/24,3\pi/24],[3\pi/24,4\pi/24],[4\pi/24,5\pi/24],[5\pi/24,\pi/4]$$
The midpoints of these sub intervals are
$$\{\frac{\pi}{48}, \frac{3 \pi}{48}, \ldots, \frac{11 \pi}{48}\}$$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{6}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_5)\right]\Delta x\\
&=\frac{\pi}{12}\left[ f( \pi/48)+ f(3\pi/48)+\cdots + f(11\pi/48)\right]\\
&\approx 0.88
\end{align*}
Now find $T_6$
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{6}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+2f(x_4)+2f(x_5)+f(x_6)\right]\Delta x\\
&=\frac{\pi}{48}\left[f(0)+2f(\pi/24)+\cdots +2f(5\pi/24)+f(\pi/4)\right] \\
&\approx 0.883
\end{align*}
