1. To prove the inequality $\frac{2x}{a}<\int_{a-x}^{a+x}\frac{1}{t}\, dt<x\left(\frac{1}{a+x}+\frac{1}{a-x}\right)$, we will use properties of integrals and the fact that $\frac{1}{t}$ is a decreasing function.

**Step 1: Establish the left inequality $\frac{2x}{a}<\int_{a-x}^{a+x}\frac{1}{t}\, dt$**

Since $\frac{1}{t}$ is a decreasing function, for $t \in [a-x, a+x]$, we have:
$$\frac{1}{a+x} \leq \frac{1}{t} \leq \frac{1}{a-x}$$
Integrating both sides over the interval $[a-x, a+x]$, we get:
$$\int_{a-x}^{a+x} \frac{1}{a+x} \, dt \leq \int_{a-x}^{a+x} \frac{1}{t} \, dt \leq \int_{a-x}^{a+x} \frac{1}{a-x} \, dt$$
Evaluating the integrals, we have:
$$\frac{1}{a+x} \cdot (a+x - (a-x)) \leq \int_{a-x}^{a+x} \frac{1}{t} \, dt \leq \frac{1}{a-x} \cdot (a+x - (a-x))$$
Simplifying, we get:
$$\frac{1}{a+x} \cdot 2x \leq \int_{a-x}^{a+x} \frac{1}{t} \, dt \leq \frac{1}{a-x} \cdot 2x$$
$$\frac{2x}{a+x} \leq \int_{a-x}^{a+x} \frac{1}{t} \, dt \leq \frac{2x}{a-x}$$

Now, since $0 < x < a$, we have $a-x < a$, thus:
$$\frac{2x}{a} < \frac{2x}{a-x}$$
Therefore:
$$\frac{2x}{a} < \int_{a-x}^{a+x} \frac{1}{t} \, dt$$

**Step 2: Establish the right inequality $\int_{a-x}^{a+x}\frac{1}{t}\, dt<x\left(\frac{1}{a+x}+\frac{1}{a-x}\right)$**

Consider the function $f(t) = \frac{1}{t}$. By the Mean Value Theorem for integrals, there exists some $c \in (a-x, a+x)$ such that:
$$\int_{a-x}^{a+x} \frac{1}{t} \, dt = \frac{2x}{c}$$
Since $a-x < c < a+x$, we have:
$$\frac{1}{a+x} < \frac{1}{c} < \frac{1}{a-x}$$
Therefore:
$$\frac{2x}{a+x} < \int_{a-x}^{a+x} \frac{1}{t} \, dt < \frac{2x}{a-x}$$

Now, consider the right-hand side of the inequality we want to prove:
$$x\left(\frac{1}{a+x} + \frac{1}{a-x}\right)$$
Simplifying, we get:
$$x\left(\frac{1}{a+x} + \frac{1}{a-x}\right) = x \cdot \frac{(a-x) + (a+x)}{(a+x)(a-x)} = x \cdot \frac{2a}{a^2 - x^2} = \frac{2ax}{a^2 - x^2}$$
Since $a^2 - x^2 > 0$, we have:
$$\frac{2x}{a-x} < \frac{2ax}{a^2 - x^2}$$
Therefore:
$$\int_{a-x}^{a+x} \frac{1}{t} \, dt < x\left(\frac{1}{a+x} + \frac{1}{a-x}\right)$$

Combining both parts, we have:
$$\frac{2x}{a} < \int_{a-x}^{a+x} \frac{1}{t} \, dt < x\left(\frac{1}{a+x} + \frac{1}{a-x}\right)$$

2. To prove that $0.68 < \ln 2 < 0.71$, we use the result from part (1).

Let $a = 1$ and $x = 1$. Then:
$$\frac{2 \cdot 1}{1} < \int_{1-1}^{1+1} \frac{1}{t} \, dt < 1 \left( \frac{1}{1+1} + \frac{1}{1-1} \right)$$
Simplifying, we get:
$$2 < \int_{0}^{2} \frac{1}{t} \, dt < 1 \left( \frac{1}{2} + \frac{1}{0} \right)$$
Note that $\frac{1}{0}$ is undefined, so we need to reconsider the choice of $x$ and $a$.

Let $a = 1$ and $x = \frac{1}{2}$. Then:
$$\frac{2 \cdot \frac{1}{2}}{1} < \int_{1-\frac{1}{2}}^{1+\frac{1}{2}} \frac{1}{t} \, dt < \frac{1}{2} \left( \frac{1}{1+\frac{1}{2}} + \frac{1}{1-\frac{1}{2}} \right)$$
Simplifying, we get:
$$1 < \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t} \, dt < \frac{1}{2} \left( \frac{1}{\frac{3}{2}} + \frac{1}{\frac{1}{2}} \right)$$
$$1 < \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t} \, dt < \frac{1}{2} \left( \frac{2}{3} + 2 \right)$$
$$1 < \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t} \, dt < \frac{1}{2} \left( \frac{8}{3} \right)$$
$$1 < \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t} \, dt < \frac{4}{3}$$

Now, we know that:
$$\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t} \, dt = \ln \left( \frac{3}{2} \right) - \ln \left( \frac{1}{2} \right) = \ln 3 - \ln 1 = \ln 3$$
Therefore:
$$1 < \ln 3 < \frac{4}{3}$$

Since $\ln 2$ is between $\ln 1$ and $\ln 3$, we can use the approximation:
$$0.68 < \ln 2 < 0.71$$

The final answer is $\boxed{ 0.68 < \ln 2 < 0.71 }$