Let g(x) = |x² - 4 |x| + 3| for all real x.

The curve. For $x \ge 0$, $g(x) = |(x - 1)(x - 3)|$: zero at $1$ and $3$, the reflected arch between them peaks at $g(2) = 1$, and $g(0) = 3$. For $x < 0$ the curve is the mirror image. So it has four zeros, $\pm1$ and $\pm3$, two arches of height $1$, and a local peak of height $3$ at $x = 0$.

(a) $y = \tfrac12$ is below both arches' tops: on $x \ge 0$ it meets the curve once in $(0, 1)$, twice on the arch and once beyond 3. Mirrored: 8 solutions.

(b) $y = 1$ touches the tops of the arches and cuts the outer parts: for $x \ge 0$ the solutions are $2 - \sqrt{2}, 2, \sqrt{2} + 2$, and each has its mirror image, 6 solutions.

(c) $y = 3$: $x^2 - 4x + 3 = 3$ gives $x = 0$ or $x = 4$, and $x^2 - 4x + 3 = -3$ has no real root. So $x = -4, 0, 4$: 3 solutions (the point $x = 0$ counts once).

(d) $g(g(x)) = 0$ needs $g(x) \in \{1, 3\}$ (the zeros $-1$ and $-3$ cannot occur, as $g \ge 0$). From (b) and (c): $6 + 3 = 9$ solutions.

(e) $g(x) = x$ needs $x \ge 0$. Then $x^2 - 4x + 3 = x$ gives $x^2 - 5x + 3 = 0$, with roots $\frac{5}{2} - \frac{\sqrt{13}}{2} \text{ and } \frac{\sqrt{13}}{2} + \frac{5}{2}$, both positive and both outside $(1, 3)$, where $x^2 - 4x + 3 \ge 0$ as needed. And $x^2 - 4x + 3 = -x$ gives $x^2 - 3x + 3 = 0$, with no real root. So the sum is $5$.