The solution can be derived as follows:

1. Initial Capacitance (with air as the dielectric):
   For a parallel plate capacitor with air as the dielectric, the capacitance $C_{\text{initial}}$ 

   is given by: 

   $$C_{\text{initial}} = C$$The battery applies a potential difference

   $E$, so the initial charge on the capacitor is:

    

   $$Q_{\text{initial}} = C_{\text{initial}} \times E = C \times E$$ 

2. Capacitance with Dielectric Slab:
   When a dielectric slab of dielectric constant $K$ 

   is inserted between the plates, the capacitance increases by a factor of $K$ 

   , so the new capacitance $C_{\text{new}}$ 

   becomes: 

   $$C_{\text{new}} = K \times C$$ 

3. Final Charge on Capacitor:
   The battery remains connected and maintains a constant voltage $E$ 

   . Therefore, the final charge on the capacitor is: 

   $$Q_{\text{final}} = C_{\text{new}} \times E = K \times C \times E$$ 

4. Additional Charge Drawn from the Battery:
   The additional charge drawn from the battery is the difference between the final charge and the initial charge: 

   $$\Delta Q = Q_{\text{final}} - Q_{\text{initial}} = (K \times C \times E) - (C \times E)$$Simplifying:

    

   $$\Delta Q = (K - 1) \times C \times E$$ 

Thus, the additional charge drawn from the battery is:

$$(K - 1) \times C \times E$$

The capacitance of a capacitor is proportional to the permittivity of the medium. Thus, when air is replaced by a dielectric of constant K, capacity increases K times.

For an isolated capacitor, force is \(F = \frac{Q^2}{2 K A \varepsilon_0}\) which decreases with dielectric (\(K > 1\)). The electric field \(E = \frac{E_0}{K}\) also decreases. Hence, Reason is false.

When the battery is disconnected, the charge \(Q\) remains constant. The initial energy is \(U_i = \frac{Q^2}{2C_0}\). After inserting the dielectric of constant \(K\), the capacitance becomes \(C = K C_0\) and the final energy is \(U_f = \frac{Q^2}{2K C_0} = \frac{U_i}{K}\). Therefore, the ratio \(U_i / U_f = K\).

The shift in plate distance to maintain constant potential is given by \(\Delta d = t\left(1 - \frac{1}{K}\right)\) where \(t\) is thickness and \(K\) is dielectric constant. Substituting the values: \(1.6 = 2\left(1 - \frac{1}{K}\right)⇒ 0.8 = 1 - \frac{1}{K} ⇒ K = 5\).

As the capacitor remains connected to the battery, the potential difference \(V\) remains constant. The stored energy is \(U = \frac{1}{2} C V^2\). Since slipping a dielectric slab increases the capacitance \(C\), the energy increases correspondingly.

Once disconnected, charge \(Q\) remains constant. Introducing a slab increases capacitance \(C\). Since potential difference \(V = Q/C\) and stored energy \(U = \frac{Q^2}{2C}\), both potential difference and stored energy decrease.

The initial energy of the isolated capacitor is \(U_i = \frac{1}{2} C V^2 = \frac{\varepsilon_0 A V^2}{2d}\). After the dielectric is introduced, capacitance becomes \(kC\) and energy becomes \(U_f = \frac{U_i}{k}\). The work done by the system is \(-\Delta U = U_i - U_f = \frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1 - \frac{1}{k}\right)\).

Using conservation of charge, the initial charge is \(Q = C V_1\). When connected in parallel to an uncharged capacitor of capacitance \(kC\), the total capacitance becomes \(C(1+k)\). Thus, \(C V_1 = C(1+k)V_2 implies k = \frac{V_1-V_2}{V_2}\).

Initial capacitance \(C_0 = \frac{\varepsilon_0 A}{d} = 1\text{ pF}\). Doubling the distance and inserting a dielectric \(k\) makes the capacitance \(C = \frac{k \varepsilon_0 A}{2d} = \frac{k}{2} C_0\). Since \(C = 2\text{ pF}\), we obtain \(2 = \frac{k}{2}(1) ⇒ k = 4\).