(i) The capacitance of the capacitor will rise as the dielectric slab is inserted between its of plates. As a result, the capacitor's potential drop will be smaller (V=‌

Q

C

). As a result (since, both are connected in series), the potential drop across the bulb will rise. Its brightness will consequently rise.
(ii) The potential drop across the resistor will rise as resistance (R) is increased. The potential drop over the bulb will therefore be less (since, both are connected in series). So, it will become dimmer.