(a)(i) By definition, electric field at a point is force per unit test charge: $\vec E=\dfrac{\vec F}{q}$, i.e. $\vec F=q\vec E$.

(ii) Potential at a point due to charge $Q$: work done in bringing unit positive charge from infinity, $V=\dfrac{W}{q}=\dfrac{kQ}{r}$; and $E=-\dfrac{dV}{dr}$ relates the two.

(b) Three charges $q$ at the vertices of an equilateral triangle of side $a$: force between any pair, $F=\dfrac{kq^2}{a^2}$. The net force on one charge due to the other two is the vector sum of two such forces at $60°$ to each other (interior angle of the triangle):

$$F_{net}=\sqrt{F^2+F^2+2F^2\cos60°}=\sqrt{2F^2(1+0.5)}=F\sqrt3=\frac{\sqrt3\,kq^2}{a^2}$$

directed outward along the median from that vertex (bisecting the angle between the two individual forces).

(a) Energy stored in air-filled parallel plate capacitor: work done in charging from 0 to $Q$:

$$U=\int_0^Q\frac{q}{C}dq=\frac{Q^2}{2C}=\frac12CV^2$$

With dielectric $K$ inserted (charge $Q$ constant, isolated capacitor): $C\to KC$, so $U'=\dfrac{Q^2}{2KC}=\dfrac{U}{K}$ — energy DECREASES by factor $K$.

(b) Two dielectric slabs $K_1$, $K_2$ each filling half the plate area $A/2$ (full separation $d$) act as two capacitors in PARALLEL:

$$C_1=\frac{K_1\varepsilon_0(A/2)}{d},\quad C_2=\frac{K_2\varepsilon_0(A/2)}{d}$$

$$C_{eff}=C_1+C_2=\frac{\varepsilon_0A(K_1+K_2)}{2d}$$