Burned out filament but the light bulb has small resistance
These bulbs contain a shunt wire which is normally insulated from the bulb but closes when a high voltage is applied. From How Stuff Works:



You can also see the wire in your image, below the burned out filament.

The shunt wire is coated in an insulation with a low breakdown voltage. When a bulb burns out, the other bulbs look like wires (especially after they cool and their resistance decreases) and the burnt bulb looks like an open circuit, thus nearly the full supply voltage (110 or 220 VAC) appears across the broken bulb. This arcs through the shunt wire's insulation, destroying the insulation, allowing the shunt wire to short out the bulb, allowing the string to continue working.







simulate this circuit &ndash; Schematic created using CircuitLab

This is an example of an antifuse. An ordinary fuse starts with a low resistance, and converts to a high resistance above a certain current threshold. An antifuse does the opposite in every respect: it starts with a high resistance, and converts to a low resistance above a certain voltage threshold. The two could be considered electrical duals.

Those Christmas light fixers that look like guns generate a high voltage spike which can help a failed antifuse work. They very briefly generate a voltage higher than mains voltage (not unlike a camera flash) which can help the antifuse do its thing if mains voltage isn't enough to break down the insulation.