Circuit question (physics)?

Resistance of a₁, R₁ = Uₒ²/P₁ = Uₒ²/40 Ω
Resistance of a₂, R₂ = Uₒ²/P₂ = Uₒ²/60 Ω
Equivalent resistance of the two lamps, R = (Uₒ²/40) + (Uₒ²/60) = 0.416Uₒ²

Current of the series current, I = Uₒ/R = Uₒ/(0.416Uₒ²) = 24/Uₒ A

In the series circuit:
Power of a₁ = I²R₁ = (24/Uₒ)² × (Uₒ/40) = 14.4 W
Power of a₂ = I²R₂ = (24/Uₒ)² × (Uₒ²/60) = 9.6 W

a₁ will shine brighter because of the higher power.

filament lamps are very non-linear vs voltage, so this cannot be solved.

BUT, if I assume they are linear...

power = voltage²/resistance
for the first, 40 = U₀²/R₁
for the second, 60 = U₀²/R₂

R₁ = U₀²/40
R₂ = U₀²/60
note that R₂ < R₁

now put them in series. Use voltage divider rule.
U₁ = U₀R₁/(R₁+R₂)
U₂ = U₀R₂/(R₁+R₂)

power = U²/R
P₁ = (U₀R₁/(R₁+R₂))²/R₁
P₂ = (U₀R₂/(R₁+R₂))²/R₂

P₁ = R₁(U₀/(R₁+R₂))²
P₂ = R₂(U₀/(R₁+R₂))²

 now compare the two. 
(U₀/(R₁+R₂))² term is same, so power is proportional to resistance
since R₂ < R₁, that means P₂ < P₁ and A₁ is brightest

R1 = Uo^2/40
R2 = Uo^2/60
I = Uo/(R1+R2) = Uo*120/5Uo^2 = 24/Uo
P1 = R1*I^2 = Uo^2/40*24^2/Uo^2 = 14,4 watt
P2 = R2*I^2 = Uo^2/60*24^2/Uo^2 = 9,6 watt 
total P = 24 watt
or more simply : 
equivalent power in series is the parallel of the powers, i.e. 60 // 40 = 24 watt
P1 = P*a1/(a1+a2) = 24*60/100 = 14,4 watt
P2 = Pa2/(a1+a2) = 24*40/100 = 9,6 watt

a1 will shine more than a2

P = V²/R so R = V²/P = Uo²/P
The current i = Uo/(Uo²/40 + Uo²/60) = 
Uo/(60Uo²/2400 + 40Uo²/2400) = 2400Uo/Uo²(60+40)
= 24/Uo = i in the ckt with the two resistors in series with source Uo. Thus the power in the 40W
= 24²/Uo² *Uo²/40 = 24²/40 = 14.4W
In the 60W P = 24²/60 = 9.6W so the 40W bulb is brighter.

Each lamp will have 1/2 design voltage.
The 40W a1 will dissipate ~10W.
The 60W a2 will dissipate ~15W.
The 60W bulb a2 will be brighter.