Denote by $C(O, R)$ the circle with center $O$ and radius $R$. If $\omega_1 = C(O_1, R_1)$ and $\omega_2 = C(O_2, R_2)$, then
$$d(\omega_1, \omega_2)^2 = O_1 O_2^2 - (R_1 - R_2)^2,$$
and $d(\omega_1, \omega_2)$ is defined if and only if the right hand side is nonnegative. This is equivalent to the existence of the common external tangents of these circles.

a) Let $\omega = C(O, R)$ and $\omega_i = C(O_i, R_i)$ for $1 \leq i \leq n$. We have
$$\frac{1}{n} \sum_{i=1}^{n} d(\omega, \omega_i)^2 = \frac{1}{n} \sum_{i=1}^{n} OO_i^2 + \frac{1}{n} \sum_{i=1}^{n} (R - R_i)^2.$$
Suppose that $\bar{R} = \frac{1}{n}(R_1 + R_2 + \dots + R_n)$. Then
$$\frac{1}{n} \sum_{i=1}^{n} (R - R_i)^2 = R^2 - 2R\bar{R} + \frac{1}{n} \sum_{i=1}^{n} R_i^2 = (R - \bar{R})^2 + \frac{1}{n} \sum_{i=1}^{n} (\bar{R} - R_i)^2.$$
Similarly, let $\bar{O}$ be the centroid of all the $O_i$'s ($1 \leq i \leq n$). Then
$$\frac{1}{n} \sum_{i=1}^{n} OO_i^2 = O\bar{O}^2 + \frac{1}{n} \sum_{i=1}^{n} \bar{O}O_i^2.$$
Therefore, the circle $C(\bar{O}, \bar{R})$ satisfies the problems condition. Now, if two circles $\bar{\omega}_1 = C(P_1, r_1)$ and $\bar{\omega}_2 = C(P_2, r_2)$ both satisfy the problems condition, for each circle $\omega$ we can write (C is a constant value here):
$$\begin{align*} d(\omega, \bar{\omega}_1)^2 - d(\omega, \bar{\omega}_2)^2 &= C \\ \Rightarrow \quad &OP_1^2 - OP_2^2 + (R - r_1)^2 - (R - r_2)^2 = C \\ \Rightarrow \quad &OP_1^2 - OP_2^2 - 2R(r_1 - r_2) = C \end{align*}$$
By fixing $O$, we have $r_1 = r_2$ and hence $OP_1^2 - OP_2^2$ is a constant value. Thus $P_1 = P_2$ and $\bar{\omega}_1 = \bar{\omega}_2$. Therefore, $\bar{\omega} = C(\bar{O}, \bar{R})$ is the unique circle satisfying the problems condition.

b) We can write $x_3 = (1 - \alpha)x_1 + \alpha x_2$ for some $\alpha \in \mathbb{R}$. Since the distances of $O_1$ and $O_2$ from the common tangent are $R_1$ and $R_2$, respectively, we conclude that the distance of $O_3$ from this line is $|(1 - \alpha)R_1 + \alpha R_2|$ (If the number in the absolute value sign is negative, it means that $O_1$ and $O_2$ are in the two sides of that line). So
$$R_3 = |(1 - \alpha)R_1 + \alpha R_2|.$$
Furthermore, the centroid of $\omega_1$ and $\omega_2$ is $\bar{\omega} = C(\frac{x_1+x_2}{2}, \frac{R_1+R_2}{2})$. Therefore, putting $\alpha = \frac{1}{2}$ in the definition of $\omega_3$ implies
$$\begin{aligned} d(\omega, \omega_1) &= d(\omega, \omega_2) \\ &\Rightarrow OO_1^2 - (R - R_1)^2 = OO_2^2 - (R - R_2)^2 \\ &\Rightarrow OO_1^2 - OO_2^2 = R_1^2 - R_2^2 - 2R(R_1 - R_2) \\ &\Rightarrow (x - x_1)^2 - (x - x_2)^2 = R_1^2 - R_2^2 - 2R(R_1 - R_2) \\ &\Rightarrow 2x(x_1 - x_2) - 2R(R_1 - R_2) = x_1^2 - x_2^2 - R_1^2 + R_2^2. \end{aligned}$$
We have
$$\begin{aligned} d(\omega, \omega_3)^2 &= (x - x_3)^2 + y^2 - (R - R_3)^2 \\ &= (x - ((1 - \alpha)x_1 + \alpha x_2))^2 + y^2 - (R - ((1 - \alpha)R_1 + \alpha R_2))^2 \\ &= ((x - x_1) + \alpha(x_1 - x_2))^2 + y^2 - ((R - R_1) + \alpha(R_1 - R_2))^2. \end{aligned}$$
Suppose that circles $\omega$, $\omega_1$ and $\omega_2$ are fixed and $\alpha$ varies. We can write $d(\omega, \omega_3)^2$ as a polynomial of $\alpha$ such as $p_2\alpha^2 + p_1\alpha + p_0$. We know $p_2 = (x_1 - x_2)^2 - (R_1 - R_2)^2$ is nonnegative because $d(\omega_1, \omega_2)$ is defined. Furthermore, the coefficient of $\alpha$ is
$$p_1 = 2(x_1 - x_2)(x - x_1) - 2(R_1 - R_2)(R - R_1).$$
Because of the relation between $X$ and $R$,
$$x_1^2 - R_1^2 - x_2^2 + R_2^2 - 2x_1(x_1 - x_2) + 2R_1(R_1 - R_2) = -(x_1 - x_2)^2 + (R_1 - R_2)^2.$$
If $p_1 = p_2 = 0$, $d(\omega, \omega_3)$ is independent of $\alpha$. Otherwise, the minimum value of $d(\omega, \omega_3)$ is for $\alpha = \frac{1}{2}$ and the assertion is proved.

c) First, note that the radical center of three circles $\omega_1, \omega_2$ and $\omega_3$, as a circle with radius zero is equidistant from these circles (if this point is outside of all the circles). Therefore, we have
Lemma 1. If two circles $C_1$ and $C_2$ are both equidistant from $\omega_1$ and $\omega_2$, then the direct homothetic center of $C_1$ and $C_2$ lies on the radical axis of $\omega_1$ and $\omega_2$. If two circles $C_1$ and $C_2$ have equal radius then the direct homothetic center of them is not defined. In this case, the line passing through the centers of $C_1$ and $C_2$ is parallel to the radical axis of $\omega_1$ and $\omega_2$.
Let $\omega_i = C(O_i, R_i)$ and $C_i = C(P_i, r_i)$. We can assume that $O_i = (x_i, 0)$ and $P_i = (a_i, b_i)$. If $(x, 0)$ lies on the radical axis of $\omega_1$ and $\omega_2$, then
$$(x - x_1)^2 - R_1^2 = (x - x_2)^2 - R_2^2 \Rightarrow x = \frac{x_1^2 - x_2^2 - R_1^2 + R_2^2}{2(x_1 - x_2)}.$$
Denote this value by $c$. We proved in part b that
$$2a_i(x_1 - x_2) - 2r_i(R_1 - R_2) = x_1^2 - R_1^2 - x_2^2 + R_2^2 \Rightarrow a_i = r_i \frac{R_1 - R_2}{x_1 - x_2} + c.$$
The direct homothetic center of $C_1$ and $C_2$ lies on $P_1P_2$ and hence it can be shown that
$$S = \frac{r_2}{r_2 - r_1} P_1 - \frac{r_1}{r_2 - r_1} P_2 \text{ (whenever } r_1 \neq r_2),$$
so the first coordinate of $S$ is
$$\frac{r_2 a_1 - r_1 a_2}{r_2 - r_1} = \frac{1}{r_2 - r_1} \left[ r_2 \left( r_1 \frac{R_1 - R_2}{x_1 - x_2} + c \right) - r_1 \left( r_2 \frac{R_1 - R_2}{x_1 - x_2} + c \right) \right] = c.$$
Hence $S$ lies on the radical axis of $\omega_1$ and $\omega_2$.
Otherwise, if $r_1 = r_2$, by the above equations we get $a_1 = a_2$ and thus the line passing through the centers of $C_1$ and $C_2$ is parallel to the radical axis of $\omega_1$ and $\omega_2$.

d) The answer is no. Let $\omega_i = C(O_i, R_i)$ and $d_{ij} = |O_i - O_j|$. We assume that $R_1 \ge R_2 \ge R_3 \ge R_4$. According to the problems assumption we must have
$$d_{ij}^2 - (R_i - R_j)^2 = 1.$$
Note that these equations are invariant under addition of a constant number to $R_i$'s ($1 \le i \le n$). Therefore, we can suppose that $R_4 = 0$ and $O_4$ lies on the radical axis of $\omega_1$ and $\omega_2$. According to the following figure we have:
$$\begin{aligned} a^2 + x^2 - R_1^2 &= 1, \\ a^2 + y^2 - R_2^2 &= 1, \\ (x + y)^2 - (R_1 - R_2)^2 &= 1. \end{aligned}$$

Subtracting the sum of the first two equations from the third one implies:
$$2xy + 2R_1R_2 + 2 - 2a^2 = 1 \Rightarrow a = \sqrt{\frac{1}{2} + xy + R_1R_2}.$$
On the other hand, by subtraction of the first equation from the second one we have:
$$\begin{cases} x^2 - y^2 = R_1^2 - R_2^2 \\ x + y = d_{12} \end{cases} \Rightarrow \begin{aligned} x - y &= \frac{R_1^2 - R_2^2}{d_{12}} \\ \Rightarrow \{x, y\} = \frac{d_{12}}{2} \pm \frac{R_1^2 - R_2^2}{2d_{12}} \\ \Rightarrow xy = \frac{d_{12}^2}{4} - \left(\frac{R_1^2 - R_2^2}{d_{12}}\right)^2. \end{aligned}$$
Hence
$$a = \sqrt{\frac{1}{2} + \frac{d_{12}^2}{4} - \frac{(R_1^2 - R_2^2)^2}{4d_{12}^2} + R_1 R_2}.$$
Now, replacing $R_1$, $R_2$ and $R_3$ by $R_1 - R_3$, $R_2 - R_3$ and 0 yields:
$$b = \sqrt{\frac{1}{2} + \frac{d_{12}^2}{4} - \frac{((R_1 - R_3)^2 - (R_2 - R_3)^2)^2}{4d_{12}^2} + (R_1 - R_3)(R_2 - R_3)}.$$

There are two possibilities.
Case 1. $O_3$ and $O_4$ are not on the same side of $O_1O_2$. In this case we have
$$O_3O_4^2 - R_3^2 = 1 \Rightarrow 1 + R_3^2 = O_3O_4^2 \geq (a+b)^2 \geq a^2 + b^2.$$
On the other hand,
$$a^2 > \frac{1}{2} + R_1 R_2,$$
$$b^2 > \frac{1}{2} + (R_1 - R_3)(R_2 - R_3),$$
so
$$R_3^2 > R_1 R_2 + (R_1 - R_3)(R_2 - R_3) \Rightarrow (R_1 + R_2)R_3 > 2R_1 R_2.$$
But this is in contradiction with the assumption that $R_1 \geq R_2 \geq R_3$.

Case 2. $O_3$ and $O_4$ are on the same side of $O_1O_2$. In this case we have
$$\begin{aligned} 1 + R_3^2 &= (a-b)^2 + (x-z)^2 = (a^2+x^2) + (b^2+z^2) - 2ab - 2xz \\ &= 1 + R_1^2 + 1 + (R_1 - R_3)^2 - 2ab - 2xz \\ \Rightarrow 0 &= 1 + 2R_1^2 - 2R_1R_3 - 2ab - 2xz \\ \Rightarrow 4ab + 4xz &= 2 + 4R_1(R_1 - R_3). \end{aligned}$$
Substituting $d_{12}^2 = 1 + (R_1 - R_2)^2$ implies
$$a = \sqrt{\frac{1}{2} + \frac{d_{12}^2}{4} - \frac{(R_1^2 - R_2^2)^2}{4d_{12}^2} + R_1 R_2} = \sqrt{\frac{3}{4} + \frac{(R_1 + R_2)^2}{4} - \frac{(R_1 - R_2)^2 (R_1 + R_2)^2}{4(1 + (R_1 - R_2)^2)}}$$
Similarly,
$$b = \sqrt{\frac{3}{4} + \frac{(R_1 + R_2 - 2R_3)^2}{4} - \frac{(R_1 - R_2)^2 (R_1 + R_2 - 2R_3)^2}{4(1 + (R_1 - R_2)^2)}}$$
Defining
$$\begin{aligned} t &:= R_1 - R_2, \\ s &:= R_1 + R_2 - 2R_3, \\ r &:= R_1 + R_2, \end{aligned}$$
we have
$$\begin{aligned} a &= \sqrt{\frac{3}{4} + \frac{r^2}{4} - \frac{r^2 t^2}{4(1+t^2)}} = \sqrt{\frac{3}{4} + \frac{r^2}{4(1+t^2)}}, \\ b &= \sqrt{\frac{3}{4} + \frac{s^2}{4(1+t^2)}}. \end{aligned}$$
Also,
$$xz = \left(\frac{d_{12}}{2} + \frac{R_1^2 - R_2^2}{2d_{12}}\right) \left(\frac{d_{12}}{2} + \frac{(R_1 - R_3)^2 - (R_2 - R_3)^2}{2d_{12}}\right) = \frac{1+t^2}{4} + \frac{rt}{4} + \frac{st}{4} + \frac{rst^2}{4(1+t^2)}.$$
Now, we have:
$$\begin{aligned} 4ab + 4xz &= 2 + 4R_1(R_1 - R_3) \\ \Rightarrow \sqrt{\left(3 + \frac{r^2}{1+t^2}\right) \left(3 + \frac{s^2}{1+t^2}\right)} + 1 + t^2 + rt + st + \frac{rst^2}{1+t^2} &= 2 + (r+t)(s+t). \end{aligned}$$
Multiplication by $1+t^2$ and some computations yields:
$$\sqrt{(3 + 3t^2 + r^2)(3 + 3t^2 + s^2)} = 1 + t^2 + rs,$$
which is in contradiction with the Cauchy-Schwarz inequality.