$\left(a\right)$ Ionic character

$\%ioniccharacter=16(x_A-x_B)+3.5{(x_A-x_B)}^2$

where, $x_A-x_B=$Difference in electro-negativity of bonded atoms.$16\times (3.0-0.7)+3.5\times {(3.0-0.7)}^2=55.32\%$

$\left(b\right)$ Geometry for both the species is tetrahedral, as in both 'O' is $s{p}^3$ hybridised

$\left(c\right) 113=\left[Rn\right] 5f^{14}6d^{10}7s^{2}7p^1=$ Due to inert pair effect $7s^2$electron do not take part in bond formation, so, most stable oxidation state is $+1.$

$\left(d\right)$ Successive ionization energies are higher for every element. $2^{nd}$ ionization of K happens from $3s^{2}3p^6$ configuration which is an inert gas configuration and for $Ca$ it happens from $4s^1$. So, K has higher second ionisation energy than Ca.