Solution:

a.
Extend $B I$ to meet the circumcircle in $F$. Then we know that $F A = F I = F C$. (See Figure)

Let $B I : I F = \lambda : \mu$. Applying Stewart's theorem to triangle $B A F$, we get
$$\lambda A F^{2} + \mu A B^{2} = (\lambda + \mu)\left(A I^{2} + B I \cdot I F\right)$$
Similarly, Stewart's theorem to triangle $B C F$ gives
$$\lambda C F^{2} + \mu B C^{2} = (\lambda + \mu)\left(C I^{2} + B I \cdot I F\right)$$
Since $C F = A F$, subtraction gives
$$\mu\left(A B^{2} - B C^{2}\right) = (\lambda + \mu)\left(A I^{2} - C I^{2}\right)$$
Using the standard notations $A B = c$, $B C = a$, $C A = b$ and $s = (a + b + c) / 2$, we get $A I^{2} = r^{2} + (s - a)^{2}$ and $C I^{2} = r^{2} + (s - c)^{2}$ where $r$ is the in-radius of $A B C$. Thus
$$\mu\left(c^{2} - a^{2}\right) = (\lambda + \mu)\left((s - a)^{2} - (s - c)^{2}\right) = (\lambda + \mu)(c - a) b$$
It follows that either $c = a$ or $\mu(c + a) = (\lambda + \mu) b$. But $c = a$ implies that $a = b = c$ since $a, b, c$ are in arithmetic progression. However, we have taken a non-equilateral triangle $A B C$. Thus $c \neq a$ and we have $\mu(c + a) = (\lambda + \mu) b$. But $c + a = 2 b$ and we obtain
$$2 b \mu = (\lambda + \mu) b.$$
We conclude that $\lambda = \mu$. This in turn tells that $I$ is the mid-point of $B F$. Since $O F = O B$, we conclude that $O I$ is perpendicular to $B F$.

Alternatively

Applying Ptolemy's theorem to the cyclic quadrilateral $A B C F$, we get
$$A B \cdot C F + A F \cdot B C = B F \cdot C A$$
Since $C F = A F$, we get $C F(c + a) = B F \cdot b = B F(c + a) / 2$. This gives $B F = 2 C F = 2 I F$. Hence $I$ is the mid-point of $B F$ and as earlier we conclude that $O I$ is perpendicular to $B F$.

Alternatively

Join $B O$. We have to prove that $\angle B I O = 90^{\circ}$, which is equivalent to $B I^{2} + I O^{2} = B O^{2}$. Draw $I L$ perpendicular to $A B$. Let $R$ denote the circumradius of $A B C$ and let $\triangle$ denote its area. Observe that $B O = R$, $I O^{2} = R^{2} - 2 R r$,
$$B I = \frac{B L}{\cos (B / 2)} = (s - b) \sqrt{\frac{c a}{s(s - b)}}$$
Thus we obtain
$$B I^{2} = a c (s - b) / s = \frac{a c}{3}$$
since $a, b, c$ are in arithmetic progression. Thus we need to prove that
$$\frac{a c}{3} + R^{2} - 2 R r = R^{2}$$
This reduces to proving $2 R r = a c / 3$. But
$$2 R r = 2 \cdot \frac{a b c}{4 \Delta} \cdot \frac{\Delta}{s} = \frac{a b c}{2 s} = \frac{a b c}{a + b + c} = \frac{a c}{3}$$
using $a + c = 2 b$. This proves the claim.

b.
Join $I D$. Note that $\angle B I O = \angle B D O = 90^{\circ}$. Hence $B, D, I, O$ are concyclic and hence $\angle B I D = \angle B O D = A$. Since $\angle D B I = \angle K B A = B / 2$, it follows that triangles $B A K$ and $B I D$ are similar. This gives
$$\frac{B A}{B I} = \frac{B K}{B D} = \frac{A K}{I D}$$
However, we have seen earlier that $B I = a c / 3$. Moreover $A K = b c / (a + c)$. Thus we obtain
$$B K = \frac{B A \cdot B D}{B I} = \frac{1}{2} \sqrt{3 a c}, \quad I D = \frac{A K \cdot B I}{B A} = \frac{1}{2} \sqrt{\frac{a c}{3}}.$$
By symmetry, we must have $I E = \frac{1}{2} \sqrt{\frac{a c}{3}}$. Finally
$$I K = \frac{b}{a + b + c} \cdot B K = \frac{1}{3} B K = \frac{1}{2} \sqrt{\frac{a c}{3}}$$
Thus $I D = I E = I K$ and $I$ is the circumcentre of $D K E$.

Alternatively

Observe that $A K = b c / (a + c) = c / 2 = A E$. Since $A I$ bisects angle $A$, we see that $A I E$ is congruent to $A I K$. This gives $I E = I K$. Similarly $C I D$ is congruent to $C I K$ giving $I D = I K$. We conclude that $I D = I K = I E$.