A capacitor and a neon lamp focused circuit problem
It seems to me that you have confused the graph description with the discharge and charge time description used in the solution. 

The chaging time is \$T_C = t_3 - t_2\$ and dischaging time is \$T_D = t_2-t_1\$

And in the solution, they use \$t_1\$ to express the charging time and \$t_2\$ to express the discharging time. 

And to be able to solve this exercise all you need is this equation 

$$V_C(t) = V_S(1 - e^{\frac{-t}{RC}})$$

Do you know this formula? 

And this equation assumes that the initial voltage across the capacitor is \$V_C = 0V\$

Thus, for example, to find the charting time fist we need to rearrange the equation and find "time". 

$$t = RC\ln\left(\frac{V_S}{V_S - V_C}\right)$$

Simple Capacitor charging and discharging question

And now to find the charging time we need to find how long does it take for the capacitor to charge from \$0V\$ to the striking voltage threshold \$V_U =77V\$. 

But our capacitor was already charged to \$V_D = 35V\$ thus, we need to subtract the time that was needed to charge the capacitor from \$0V\$ to \$V_D = 35V\$.  

$$T_C = RC\ln(\frac{V_S}{V_S - V_U})-RC\ln(\frac{V_S}{V_S-V_D})= RC\ln\left(\frac{\frac{V_S}{V_S - V_U}}{\frac{V_S}{V_S - V_D}}\right) = $$

$$ = RC\ln\left(\frac{V_S}{V_S - V_U} \times \frac{V_S - V_D}{V_S}\right)= RC\ln \left(\frac{V_S - V_D}{V_S - V_U}\right) $$

We can use a similar approach for a discharge time but we need to use this equation for discharge:  

$$V_C(t) = V_{init} \times e^{\frac{-t}{RC}} $$

Or in both cases you can use this general formula for capacitor charging/discharging phase:

$$V_C(t) = V∞ + (V_{start} - V∞) \times \left(e^{\frac{-t}{RC}}\right)$$

Where:

\$V_{start}\$  initial capacitor voltage.

\$V∞\$   steady-state final voltage.