Problems solving strategies of experts in simple DC-circuit problems
Any beginners course in electronics covers problems like this, only they use resistors with given resistance values instead of unspecified bulbs.  

Suppose the power supply is 12V and each of the resistors is 2\$\Omega\$. In the case of the open switch L3 doesn't count and we have L1 and L2 in series, that's 4\$\Omega\$ in total. You can calculate the current: 12V/4\$\Omega\$ = 3A, and the power in each resistor: (3A)\$^2\$ \$\times\$ 2\$\Omega\$ = 18W.  

If you close the switch L2 and L3 are in parallel, so their combined resistance is 2\$\Omega\$/2 = 1\$\Omega\$. That is series with L1 is 3\$\Omega\$ in total. Then the current is 12V/3\$\Omega\$ = 4A. That's the current through L1. L2 and L3 each see half of that: 2A.
Power in L1 is now (4A)\$^2\$ \$\times\$ 2\$\Omega\$ = 32W, and the power in L2 and L3 (2A)\$^2\$ \$\times\$ 2\$\Omega\$ = 8W.  

So when the switch is closed L1 will light brighter, and L2 dimmer.  

edit
I seem to have missed the word "qualitatively" in your question, so I'll give you another answer.  

If the switch is closed the total resistance decreases due to L3 which becomes parallel to L2. So That means more current and L1 will light brighter. For L2 it's not so obvious: the total current has increased, but only half of that goes through L2. So you'll have to see whether the increase is more than double the original value. It isn't, because that would only be true if L2 was shorted. So L2 will light dimmer.  

I've never seen a resource where this reasoning is explained. Logical thinking seems to be the key, maybe a course in logic (formal systems) would help. In a formal system you describe the rules of the game, like "the resistance of two lamps in parallel is half the resistance of one lamp" and "the same voltage over a lower resistance causes a higher current". From this set of rules you deduce additional, dependent rules which bring you closer to a solution.