Answer
a) See details below.
b) Start from a), $\ln 2< 1 < \ln 3 \implies e^{\ln 2} < e^1 < e^{\ln3} \implies 2 < e < 3$. The first implication is due to the fact that $f(x) = e^x$ is a monotonically increasing function on the entire real line; the second implication is due to $f(g(x))=x$ if $f(x)$ and $g(x)$ are inverse functions of each other. (See a fuller discussion below.)
Work Step by Step
a)
The left part of the compound inequality ($\ln 2 < 1$) may be proven by constructing a unit square over the interval $[1,2]$, Notice that such a unit square has area $1$ (brown area in the diagram) and its area overestimates the area under $y=1/x,x>0$ over this same interval, this area being $\color{green}{\displaystyle \int_1^2 1/x\ dx = \ln 2}$ (see the diagram).
The curve $y=1/x,1\lt x\le 2,$ is totally under the horizontal segment $y=1, 10,$ is monotonically decreasing.
We have thus shown:
$\qquad \ln 2 < 1. \qquad \text{(Ineq. 1)}$
The right part of the compound inequality ($1 < \ln 3$) may be proven by computing a right-endpoint Riemann sum to estimate $\displaystyle \int_1^3 1/x\ dx = \ln 3$, this Riemann sum computed using a partition of $[1,3]$ into $8$ congruent subintervals. We will show that such an $\color{blue}{R_8 \approx 1.01988}$ (see the diagram for 4.1.b). Furthermore, this Riemann sum, $R_8$, underestimates the area $\ln 3$ as $y=1/x,x>0$ is monotonically decreasing. Consequently,
$\qquad 1 < 1.01988 < R_8 < \ln 3 \\
\qquad 1< \ln 3 \qquad (\text{Ineq. 2})$
From Ineq. 1 and Ineq. 2, we have the desired result:
$\qquad \ln 2 < 1 < \ln 3.$
$\underline{\text{To show}\ \color{blue}{R_8 \approx 1.01988}}$
function: $f(x) = 1/x, x\gt 0$
closed interval: $[a,b]= [1,3]$
number of congruent subintervals: $8$
width of subinterval: $\Delta x = \dfrac{3-1}{8} = \dfrac{2}{8} = 0.25$
partition: $1.25 < 1.5 < 1.75 < 2.0 < 2.25 < 2.5 < 2.75 < 3$
$\begin{align*}
R_n &= \sum_{k=1}^n f\left(1 + k\cdot \Delta x \right) \cdot \Delta x \\
R_8 &= \sum_{k=1}^8 \frac{1}{1 +k\cdot (0.25)} \cdot (0.25) \\
&= \sum_{k=1}^8 \frac{1}{1 + \frac{k}{4}} \cdot \frac{1}{4} \\
&= \sum_{k=1}^8 \frac{1}{4 +k} \\
&= \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} \\
\color{blue}{R_8}\ &\color{blue}{\approx 1.01988.}
\end{align*}$
b)
This follows from 4.a) and since the exponential function $f(x)=e^x$ is monotonically increasing on the real line. Thus,
$\qquad a< x < b \implies e^a < e^x < e^b. \qquad \text{(Ineq. 3)}$
Now, from 4.a), we have
$\qquad \ln 2 < 1 < \ln 3.$
Applying Ineq. 3, we have
$\qquad e^{\ln 2} < e^1 < e^{\ln 3}$.
Since, $e^{\ln x} = x$, as $f(x)=e^x$ and $g(x) = \ln x$ are inverse functions of each other, we finally have the desired result:
$\qquad 2 < e < 3.$
