1. Use half-life relation for first order reactions

For a first order reaction, $t_{1/2} = \frac{0.693}{k}$ So, half-life is inversely proportional to rate constant: $t_{1/2} \propto \frac{1}{k}$

Given, $t_{1/2}(\text{Reaction 1}) : t_{1/2}(\text{Reaction 2}) = 5:2$ Hence, $\frac{k_1}{k_2} = \frac{t_{1/2,2}}{t_{1/2,1}} = \frac{2}{5}^{-1}?$ More carefully, $\frac{0.693/k_1}{0.693/k_2} = \frac{5}{2}$ $\frac{k_2}{k_1} = \frac{5}{2}$ $\frac{k_1}{k_2} = \frac{2}{5}$

So, $k_2 = \frac{5}{2}k_1$

2. Time for a given fraction to be completed in first order reaction

For a first order reaction, $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$

If fraction reacted is $x$, then fraction remaining is $(1-x)$. Thus, $t = \frac{2.303}{k} \log \frac{1}{1-x}$

3. Compute $t_1$ for Reaction 1

Reaction 1 is completed to $\frac{2}{3}$. So fraction remaining is $1-\frac{2}{3} = \frac{1}{3}$

Therefore, $t_1 = \frac{2.303}{k_1} \log 3$

Given $\log 3 = 0.477$, $t_1 = \frac{2.303 \times 0.477}{k_1}$

4. Compute $t_2$ for Reaction 2

Reaction 2 is completed to $\frac{4}{5}$. So fraction remaining is $1-\frac{4}{5} = \frac{1}{5}$

Therefore, $t_2 = \frac{2.303}{k_2} \log 5$

Given $\log 5 = 0.699$, $t_2 = \frac{2.303 \times 0.699}{k_2}$

5. Find the ratio $t_1:t_2$

$\frac{t_1}{t_2} = \frac{\frac{2.303\log 3}{k_1}}{\frac{2.303\log 5}{k_2}}$

$\frac{t_1}{t_2} = \frac{\log 3}{\log 5}\cdot\frac{k_2}{k_1}$

Using $\frac{k_2}{k_1} = \frac{5}{2}$ we get $\frac{t_1}{t_2} = \frac{0.477}{0.699}\times \frac{5}{2}$

First, $\frac{0.477}{0.699} \approx 0.6824$

Then, $\frac{t_1}{t_2} \approx 0.6824 \times 2.5 = 1.706$

So, $t_1:t_2 \approx 1.706:1$

The question asks for the value in the form $\underline{\hspace{1cm}}\times 10^{-1}$

Now, $1.706 = 17.06 \times 10^{-1}$

Nearest integer $= 17$.

6. Final Answer

$t_1:t_2 = 17 \times 10^{-1}$ So the required integer is $\boxed{17}$