How to approach such question of two periods and asymmetric information game?

You want to solve these types of games using the logic of backward induction and when beliefs become necessary, assume a pooling or separating equilibrium.

For example, start with the final nodes. It is clear that player 1 will choose C if no one has contributed before, and he is of the low type. He will choose N if someone has contributed before or if he is of the high type.

Knowing this, when player 2 makes her decision it is also easy to see that she will never contribute if player one contributed in the first period. If player 1 did not contribute in the first period, payer 2's decision depends on whether she thinks that player 1 is of the high type or of the low type. She would want to contribute if the belief that player 1 is of the high type is high enough otherwise she would prefer to choose N.

This is were assuming a separating or pooling equilibrium comes in handy. For a separating equilibrium, player 2 will learn player 1's type by the moment she has to make a decision, so she would choose C if player 1 is of the high type and choose N if it is of the low type (this is because she anticipates perfectly what player 1 will do in the second period since she knows his type).

Giving player 2's strategy, player one in the first period will choose to not contribute if he is of the high type (it is not profitable for him and he anticipates that 2 will realize he is of the high type and pay the public good), and player 1 will choose to contribute if he is of the low type (since he anticipates that player 2 will know he is of the low type and not contribute, so he would end up paying for the public good in the second period, but for him, it is better to get $b_1-underline{c}$ rather than $delta(b_1-underline{c})$).

We conclude that this is indeed an equilibrium because each type of player one ends up choosing a different action in the first period (separating) and every player is best responding to the other player's strategy, there is no profitable deviation. In this case, player 2 assigns a probability of 1 to player one being of the high type if he does not contribute in the first period and 0 if he does.

Let's now assume a pooling equilibrium. In this case when player 2 makes a decision she doesn't know what's player 1 type. She only knows that with probability $p$ he is of the low type and $1-p$ he is of the high type. Regardless of this uncertainty, if player one already paid for the public good, player 2's best response is N, if player one did not pay for the public good, player 2 can pay for it and get $b_2-c_2$ or not pay for it and get $b_2$ with probability $p$ or zero with probability $1-p$. Therefore, she chooses C if and only if $b_2-c_2geq pb_2$. That is if $b_2geq frac{c_2}{1-p}$ (i.e. if she values the public good enough).

Suppose the last inequality does not hold, in that case, player 2 does not pay for the public good. Given this strategy player one of the low types would prefer to choose C while player 1 of thee high type would prefer to choose N. This contradicts the assumption of a pooling equilibrium. In contrast, if the inequality above holds, a pooling equilibrium exists only if $b_1-underline{c}leq delta b_1$. That is if $b_1leqfrac{underline{c}}{1-delta}$ (if he doesn't value the public good to much so as to not be willing to wait).

I gave less detail for the pooling equilibrium, but hopefully, you can follow were the inequality comes from.

Lastly, for the last type of equilibrium you just have to realize that for player 2 if player 1 has not contributed to the public good yet if could be because he is of the high type (mass $=(1-p)$) or because he is of the low type and did not contribute (mass $=p(1-alpha)$). Therefore, it is of the high type with probability $frac{1-p}{1-alpha p}$ and of the low type with complementary probability. So for player two to want to randomize she should bee indifferent between contributing or not. Therefore $b_2-c_2=frac{p(1-alpha)}{1-alpha p}$. This equation gives you the equilibrium value of alpha. Similarly, player 1 of the low type should be indifferent between paying in the first period or with probability $beta$ get the good for free, and probability $1-beta$ pay for the good in the second period. Therefore $b_1-underline{c}=beta c_1+(1-beta)delta(b_1-underline{c})$. Which will give you the equilibrium value of $beta$.