Given: Area A, thickness of each dielectric d (they are the same), battery voltage V, and the two dielectric constants κ₁ and κ₂.  Question is to find the charge.

As reviewed in class, this is equivalent to two capacitors in series (if you trace the circuit through one, then you must continue through the other).

C_eq^-1= C₁^-1+C₂^-1= d/κ₁ε₀A + d/κ₂ε₀A = d/ε₀A ×(κ₁^-1+κ₂^-1)

C_eq= ε₀A/d ×κ₁κ₂/(κ₁+κ₂)

The charge is then given by the capacitance definition:  Q = C_eqV .  Note that the change on a series combination is equal to the charge on either one of the capacitors.  Although in this case that is less important, since the series combination was an artificial construct to deal with the combination.

Configuration is exactly the same as in question 21.  Note that there, the value of x was completely unnecessary.
We are given the area A, the plate separation d, the dielectric thickness b, the initial potential difference V, and the dielectric constant κ.

(a) Direct use of the capacitance formula  C = ε₀A/d

(b)  The answer is in 21(a) above.  Call this C_e

(c) Q = CV

(d) The capacitor was disconnected from the power supply, so the charge can't change.  Same answer as (c): Q' = Q

(e) The field in the empty space we can get from the equation for field near a charged conductor surface: E_s = σ/ε₀  = Q/ε₀A.  That works because all of the charge on the plates of a capacitor is always on the inside surface of the conductor.  OR, you can get the same result by using superposition of the field due to an infinite plane of charge: the induced charges on the two sides of the dielectric cancel each other, being of opposite sign and both in the same direction from the place that we are getting the field for.

(f) The field inside the dielectric is E_d = σ/κε₀ .  You can get this from the fact that reducing the field by a factor of κ from what would otherwise be there is exactly what a dielectric does, almost by definition.  OR from the rule that you should multiply any ε₀ by κ.

(g) Due to answer (d), the charge (on the plates) is unchanged, so the potential difference probably did change.  You could get this directly from the electric field: V = E_s(d−b) +E_db where the two terms are the voltage drop across the empty space and the dielectric.  OR you could use answer (b) for the capacitance and answer (d) for the charge in Q' = C_eV.

(h) External work will be the change in potiential energy (no minus sign).  At this point we know so much that we can just treat this as a changing capacitance with a constant charge, with no consideration for all the details.

W = U'−U = 0.5 Q²/C−0.5 Q²/C_e = 0.5 Q²(1/C−1/C_e)