a.
Assume $A = (x_a, y_a)$ and $B = (x_b, y_b)$ are two arbitrary points in the plane. The new perpendicular bisector of $A$ and $B$ comprises points $(x, y)$ satisfying
$$|x - x_b| + |y - y_b| = |x - x_a| + |y - y_a|.$$
Draw two lines from each of the points $A$ and $B$ parallel to the coordinate axes to create a rectangle $ACBD$. Based on the slope of $AB$ two cases are possible:
* The slope of $AB$ is $1$ or $-1$. In this case the perpendicular bisector of $A$ and $B$ is the union of segment $CD$ and the hatched parts in the figure below (The figure is drawn for the case where the slope is $1$. The other case is similar):

* The slope of $AB$ is not $\pm 1$. Assume that this slope is positive and greater than one. The following figure shows the perpendicular bisector of $A$ and $B$ in this case. Other cases are similar.

b.
Lemma. For every point $Z$ on line $XY$, $\frac{d(X,Z)}{d(Y,Z)} = \frac{XZ}{YZ}$.
Proof. It can be derived from Thales' Theorem easily.
Consider points $C$ and $D$ on line $AB$ such that $C$ is inside segment $AB$, $D$ is outside of it and $\frac{AC}{CB} = \frac{AD}{DB} = m$. Obviously, points satisfying these properties are unique and according to the lemma, $\frac{d(A,C)}{d(C,B)} = \frac{d(A,D)}{d(D,B)} = m$.
By definition, the intersection of the circle with center $A$ and radius $d(A, C)$ and the circle with center $B$ and radius $d(B, C)$ lies on the Apollonius circle (note that circles in this new metric are Euclidean squares with diagonals parallel to the coordinate axes). This intersection is a segment which is denoted by ML (as can be seen in the following figure). The intersection of similar circles for D (instead of C) determines another part of the Apollonius circle and is denoted by ON (as can be seen in the following figure). In what that follows, without loss of generality assume that $m > 1$.
An easy calculation reveals that the slope of segment LN is $\frac{m-1}{m+1}$. For an arbitrary point Z on this segment,
$$d(A, Z) = d(A, N) - \left| x_Z - x_N \right| - \left| y_Z - y_N \right|,$$
$$d(B, Z) = d(B, N) - \left| x_Z - x_N \right| + \left| y_Z - y_N \right|.$$
Since $\frac{|y_Z - y_N|}{|x_Z - x_N|} = \frac{m-1}{m+1}$ and $d(A, N) = md(B, N)$, it can be deduced that $d(A, Z) = md(B, Z)$. Therefore, all of the points of segment LN are on the Apollonius circle. A similar argument will show that all of the points of segment OM are on the Apollonius circle, too. For an arbitrary point P of the plane not on the boundary of OMLN, the quotient $\frac{d(A,P)}{d(B,P)}$ is greater or less than $m$. Therefore, the Apollonius circle of points $A$ and $B$ in this new metric comprises points on the boundary of quadrilateral OMLN.

c.
The goal is to find the parabola created by point $F = (0, 0)$ and line $l$ with equation $y = mx + c$. For an arbitrary point $P$ in the plane, draw two lines from $P$ parallel to the coordinate axes. These two lines intersect $l$ at two points, say $X$ and $Y$. It can be seen that $\min(PX, PY)$ is the distance from $P$ to $l$.
Without loss of generality, assume that $m \ge 1$ and $c > 0$. Therefore, the distance from a point $(x, y)$ in the plane to $l$ is equal to $|x - \frac{y-c}{m}|$ (length of the horizontal segment from $(x, y)$ to $l$). So for finding the points of the parabola the following equation must be solved,
$$|x| + |y| = \left|x - \frac{y-c}{m}\right| + \left|y - c\right| = \left|x - \frac{y-c}{m}\right|,$$
or equivalently,
$$m|x| + m|y| = |mx + c - y|.$$
For an arbitrary point $(x, y)$ in the plane above line $l$ ($y > mx + c$),
$$m|x| + m|y| \geq m|x| + |y| \geq -mx + y > y - mx - c = |y - mx - c|.$$
Therefore, all of the points of the parabola lie below $l$ ($y < mx + c$). Evaluating cases resulting from the sign of $x$ and $y$, it can be seen that the parabola is the union of the following four segments (Similar to the following figure):
$$\begin{cases} y = \frac{c}{m+1}, & \text{for } x > 0, y > 0 \\ y = \frac{c}{1-m}, & \text{for } x > 0, y < 0 \\ 2mx - (m+1)y + c = 0, & \text{for } x < 0, y > 0 \\ 2mx + (m-1)y + c = 0, & \text{for } x < 0, y < 0 \end{cases}$$

d.
Yes. To prove it, a lemma is needed.
Lemma. Let $A'$, $B'$ and $C'$ be three points on sides $BC$, $AC$ and $AB$ of triangle $ABC$, respectively, such that
$$\frac{d(B, A')}{d(C, A')} \frac{d(C, B')}{d(A, B')} \frac{d(A, C')}{d(B, C')} = 1.$$
Then lines $AA'$, $BB'$ and $CC'$ are concurrent.
Proof. According to the assumption and the lemma in part (b),
$$\frac{BA' \; CB' \; AC'}{CA' \; AB' \; BC'} = 1.$$
Therefore, due to Ceva's Theorem in Euclidean geometry these lines are concurrent.

Let $p = \frac{1}{2}(d(A, B) + d(B, C) + d(C, A))$. So $d(B, X_a) + d(A, B) = d(C, X_a) + d(A, C) = p$. Therefore, $d(B, X_a) = p - d(A, B)$ and $d(C, X_a) = p - d(A, C)$. Similar equalities for $X_b$ and $X_c$ imply
$$\frac{d(B, X_a)}{d(C, X_a)} \frac{d(C, X_b)}{d(A, X_b)} \frac{d(A, X_c)}{d(B, X_c)} = 1.$$
Now according to the lemma, it is easy to see that these lines are concurrent.

e.
No. Assume to the contrary that $S$ is an infinite set of points in the plane such that the mutual distances of its point are all perfect squares.
Lemma. *There is no infinite subset $\{P_1 = (x_1, y_1), P_2 = (x_2, y_2), \dots\}$ of $S$ for which the two sequences $\{x_i\}$ and $\{y_i\}$ are both monotone.*
Proof. Assume to the contrary that there exists an infinite set of points $P_i = (x_i, y_i)$ in $S$ such that both sequences $\{x_i\}$ and $\{y_i\}$ are increasing sequences (other cases can be proven similarly). For each natural number $j > 2$, let $a_j = \sqrt{d(P_1, P_j)} \in \mathbb{N}$ and $b_j = \sqrt{d(P_2, P_j)} \in \mathbb{N}$. As both sequences $\{x_j\}$ and $\{y_j\}$ are increasing, for every natural number $j > 2$,
$$a_j^2 = d(P_1, P_j) = d(P_1, P_2) + d(P_2, P_j) = d(P_1, P_2) + b_j^2.$$
Since sequences $a_j$ and $b_j$ are increasing, $d(P_1, P_2)$ can be written in infinitely many ways as the difference of two perfect squares $(d(P_1, P_2) = a_j^2 - b_j^2)$. This is impossible, however, and proves that $S$ cannot exist.
By a translation it can be assumed that $S$ contains the origin of the coordinate. It is claimed that the number of points of $S$ in the first coordinate quadrant is finite and because of symmetry, the number of points in each quadrant is finite. Therefore, $S$ is a finite set.
If set $\{y \ge 0 \mid \exists x \ge 0; (x, y) \in S\}$ is unbounded, then an infinite subset $\{P_1 = (x_1, y_1), P_2 = (x_2, y_2), \dots\}$ of $S$ can be found such that $y_1 < y_2 < \dots$. Therefore, a subsequence $\{P_{n_i}\}_i$ of $P_n$ can be founded such that the sequence $x_{n_i}$ is monotone, which is impossible according to the lemma. This means that the second coordinate of points of $S$ in the first coordinate quadrant is bounded. Similarly, it can be concluded that the first coordinate of such points is bounded. Therefore, there exists $M > 0$ such that all points of $S$ in the first coordinate quadrant lie in the square $[0, M] \times [0, M]$. Since each two points of $S$ have a distance of at least 1, only a finite number of points of $S$ can be in this square, which means that overall, $S$ is finite.