Mixed Cournot/Bertrand Duopoly
Each player seeks to maximize his or her revenue. As you said, in the Cournot model, the player varies the output. In the Bertrand model, the player varies the price. 

So we first solve for $Q_{2}$ as a function of $P_{2}$:

$$Q_{2} = 200 - 2P_{2} - 0.8Q_{1}$$

We then substitute $Q_{2}$ into $P_{1}$ to get:

$$P_{1} = 100 - 0.5Q_{1} - 0.4 (200 - 2P_{2} - 0.8Q_{1}) = 50 - 0.18Q_{1} + 0.8P_{2}$$

Player $1$'s maximization problem is:

$$\max_{Q_{1}} 50Q_{1} - 0.18Q_{1}^{2} + 0.8P_{2}Q_{1}$$

Which yields the first order conditions:

$$0.36Q_{1} = 50 + 0.8P_{2} \implies Q_{1}^{*} = \frac{1250}{9} + \frac{20}{9}P_{2}$$

Now we can substitute $Q_{1}^{*}$ into $Q_{2}$ to get:

$$Q_{2} = 200 - 2P_{2} - \frac{4}{5}(\frac{1250}{9} + \frac{20}{9}P_{2})$$

Now maximize $Q_{2}$ with respect to $P_{2}$. Can you take it from here?