\textbf{Resistances and rated currents} (treating each bulb as ohmic at its rated point):

$$R_1 = \dfrac{U^2}{P_1} = \dfrac{110^2}{100} = 121 \ \Omega, \quad R_2 = \dfrac{U^2}{P_2} = \dfrac{110^2}{40} = 302.5 \ \Omega,$$ $$I_1 = \dfrac{P_1}{U} = \dfrac{100}{110} = \dfrac{10}{11} \ \text{A} \approx 0.909 \ \text{A}, \quad I_2 = \dfrac{P_2}{U} = \dfrac{40}{110} = \dfrac{4}{11} \ \text{A} \approx 0.364 \ \text{A}.$$

\textbf{(1) Plain series across 220 V.} A series circuit forces the same current through both bulbs, but the two bulbs have different rated currents, so they cannot both run at rated power. Numerically the series current would be $I = 220/(R_1+R_2) = 220/423.5 \approx 0.52 \ \text{A}$, which is well above the $40$-W bulb's rated $0.364 \ \text{A}$ -- it would burn out -- while the $100$-W bulb runs below its rating and glows dimly. So the answer is \emph{no}.

\textbf{(2) Using the rheostat.} For both bulbs at rated values, each needs $110 \ \text{V}$ across it, which adds to $220 \ \text{V}$, so the two bulbs go in series. The $100$-W bulb needs the larger current $I_1 = 10/11 \ \text{A}$; the $40$-W bulb only needs $I_2 = 4/11 \ \text{A}$. The extra $I_1 - I_2 = 6/11 \ \text{A}$ must bypass the $40$-W bulb -- so connect the rheostat \emph{in parallel with the $40$-W bulb}. The rheostat then carries $6/11 \ \text{A}$ at $110 \ \text{V}$:

$$R = \dfrac{110}{6/11} = \dfrac{110 \times 11}{6} = \dfrac{1210}{6} \approx 201.7 \ \Omega.$$