All I can figure out is that the train is $4$ times faster than you.
Let's say that you can run with a speed $v$, and the train, $u$. Let's also assume that the tunnel has length $8L$, and that the train is originally a distance $x$ away from the entrance of the tunnel. Then we have $$\frac{3L}{v} = \frac{x}{u}$$ and $$\frac{5L}{v} = \frac{x+8L}{u}$$
We therefore have $$\frac{5}{3} \cdot \frac{x}{u} = \frac{x+8L}{u}$$
Solving this, we get $$\frac{2}{3} x = 8L \Rightarrow x = 12L$$
Plugging it back in, we get $$\frac{3L}{v} = \frac{12L}{u} \Rightarrow \frac{u}{v} = 4$$
Okay. Let $t_0$ be the velocity of the train. Let $x_0$ be the velocity of the person walking in the same direction of the train. Lets assume these values are positive. The velocity of the person walking in the other direction is negative because they are walking in the opposite direction. Let $-y_0$ be the velocity of that person.
Now... Newtonian relativity.... Imagine the entire situation, both people, train, and miles of train track you on a platform traveling with a constant velocity of $c$. If we ignore wind resistance and near light speed compaction nothing will change. The train will still pass everyone at the same time. So we can add a constant $c$ to $t_0, y_0,$ and $x_0$ and it will not change our answer and results.
So let $c = y_0$. Now let $t = t_0 + y_0$, $x = x_0 + y_0$ and $y = -y_0 + y_0 = 0$. This would be a situation where the second person is standing still, the first person is walking toward him with a combined speed of her "real" speed plus the second person's speed, and the train is travelling at its "real" speed plus the second guy's speed. Nothing changes. Another way of thinking of this is to imagine the referee guy who is watching and timing this is on a floating surfboard and gliding past the scene at a velocity of $y_0$.
Okay.... Now let $U$ be one unit equal to one train-length. The train passes person 2 in $18$ sec so we know that $t = \frac {1U}{18sec}$.
Now the train passes the first person in $20$ sec. The distance the first person walks in the $20$ seconds is $x\cdot 20sec$. So the distance the train traveled in passing person 1 is its own length plus $x\cdot 20sec$ or $1U + x\cdot 20sec$. But if it took $20 sec$ we can also express that distance as $20\sec \cdot \frac {1U}{18} = \frac {10}9 U$.
So $1U + x\cdot 20sec= \frac {10}9 U$
$x\cdot 20sec = \frac 19 U$
$x =\frac {1U}{180 sec}$.
Now lets fix this point and place in time. The tail of the train has just past person 1. Let's call the point where the head of the train is now $0$. Person 1 is $1U$ away and lets call that $ - 1U$. And person 2 is standing still at some point ahead. It takes the head of the train $10$ minutes $= 600$ secs to reach him so he is at $600\sec \cdot \frac {1U}{18sec} = \frac {300}9 U$.
Now let $18$ seconds pass so that the train passes person 2. So $618$ second have passed since the train passed person 1 at $-1U$. So person 1 is now at point $-1U + 618\cdot\frac {1U}{180sec}=\frac {73}{30}U$.
So the distance between Person 1 and Person 2 is now $\frac {300}9U -\frac {73}{30}U=\frac {309}{10}U$.
And now the time will take for person 2 to reach person 1 is $\frac {distance}{speed} = \frac {\frac {309}{10}U}{\frac {1U}{180sec}}=5562 sec =92 min 42 sec$
While the method is sound, we could have tried to solve it without the relativity shift but we'd have had an unsolved variable in all our equations that would have cancelled out in the end.
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You asked in comments if changing our frame of reference changes our measured distances and speeds and give a wrong answer. It does change ou measured distance and therefore our speeds but not our times.
If we set person 2's speed (velocity without direction to $c$ and the train length to $1U$ we get. Let $t = $ train speed and $x=$ person 1's speed.
It takes $18$ seconds for the train to pass person 2. In $18$ seconds at a speed of $c\frac {U}{sec}$ person 2 will have walked $18c U$. The train in the opposite direction passes him after he walks $18c U$ so the train will have traveled $1U - 18cU = (1-18c)U$. Thus the trains speed is $\frac {(1-18c)U}{18 sec}=(\frac 1{18} - c)\frac U{sec}$.
Let figure out Persons 1 speed. In $20$ seconds she would have walked $20\sec\cdot x\frac U{sec}=20 xU$ if her speed is $x\frac U{sec}$. Meanwhile the head of the train would have traveled $20\cdot (\frac 1{18} - c)= (\frac {10}9 -20c)U$. But as person 2 and the tail of the train are both at $20xU$ we have $20xU + 1U = (\frac {10}9 -20c)U$. Thus $x\frac U{sec}=(\frac 1{180} - c)\frac {U}{sec}$.
Now lets consider that we've had $20$ seconds (for the train to pass Person 1) and $10min = 600sec$ for the trains head to meet person 2. And $18$ seconds to pass Person two. The $638 sec$. Let's figure out where Person 1 is and where Person 2 and the tail of the train is.
Person 1 will have traveled $638(\frac 1{180}-c)U=(\frac{319}{90}- 638c)U$. The head of the train will have traveled $638(\frac 1{18} - c)U=(\frac {319}9 - 638c)U$ and so its tail and Person 2 will be at $(\frac {319}9 - 638c)U-1U = (\frac {310}9 - 638c)U$.
And the distance between Person 1 and person 2 would be $(\frac {310}9 - 638c)U - (\frac {319}{90} - 638c)U = \frac {3100-319}{90}U =\frac {2781}{90}U$.
The combined speed of Person 1 and 2 is $(\frac 1{180} - c)\frac U{sec} + c\frac {U}{sec} = \frac 1{180}\frac {U}{sec}$.
So the time it takes for them to meet will be $\frac {\frac {2781}{90}U}{\frac 1{180}\frac U{sec}} = 5562 sec=92 min 42 sec$.
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Note we never know how fast any one person or train was going nor any of the distance or positions (except for where person 1 and 2 start at [person 1 started at $0$. Person 2 started and $638\cdot \frac 1{180}U$]. The only thing we did know was the times when any two object pass each other and the combined speeds of Person 2 and Person 1, of Person 2 and the train, and the difference of speeds of Person 1 and the train.
Best Answer
Let say the train is traveling at $x$ meters per second and Nino walks at $y$ meters per second. Let us also say the train has length $L$ meters.
So first, we see that it takes Nino $45/y$ seconds before the train passes her. In that time the last wagon in the train has traveled $L+45$ meters. But using the speed of the train, we have that it has also traveled $45x/y$ meters. So we have
$$ 45\frac{x}{y} = L + 45.$$
Similarly in the second scenario, it takes $30/y$ seconds. But now the wagon has traveled $L-30$. So we have
$$ 30\frac{x}{y} = L - 30. $$
Using both we get
$$\frac{L+45}{45} = \frac{L-30}{30} \Rightarrow 15L = 2*45*30 \Rightarrow L = 180m.$$
Whats interesting about this is that you don't actually need the speeds!