The lamp is likely short-circuited/bypassed by a low-resistance path. [2]
A shorted lamp has nearly zero p.d. across it. Almost the full supply p.d. appears across the series resistor, and little/no power is dissipated in the lamp, so it is dark. [3]
If the lamp were open-circuit, current would be zero, the resistor p.d. would be zero, and the lamp/open gap would have approximately the full supply p.d. [3]
A high-resistance voltmeter draws negligible current and therefore minimally changes the p.d. it is measuring. [2]
Q2 Brightness comparison
Lamp in A and each lamp in C have the same brightness for an ideal battery; each lamp in B is dimmer. [3]
In A and each parallel branch of C, each lamp receives full battery p.d., so power per lamp is V2/R. In B, the p.d. is shared, so each lamp has smaller p.d. and lower power. [4]
Circuit C draws twice the current of A because there are two identical full-voltage branches. [2]