This problem counts the distinct real roots of the equation x² = k eˣ, where k is a real constant…

The shape of $f$. $f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x)e^{-x}$. So $f$ decreases on $(-\infty, 0)$, increases on $(0, 2)$ and decreases on $(2, \infty)$. It falls from $+\infty$ to its local minimum $f(0) = 0$, rises to its local maximum at $x = 2$, then falls towards $0$ without reaching it (it stays positive).

(a), (b) The local maximum is at $x = 2$, with value $f(2) = \frac{4}{e^{2}} \approx 0.5413$, that is 0.54.

(c) $k = 0.5$ lies between $0$ and $\dfrac{4}{e^2} \approx 0.5413$. The line $y = 0.5$ meets the falling left branch once (at a negative $x$), the rising part on $(0, 2)$ once and the falling part beyond $x = 2$ once: 3 roots.

(d) $k = 0.6$ is above the local maximum, so the line misses the hump and meets only the left branch: 1 root.

(e) Exactly two roots happen only when the line touches the hump: $k = \dfrac{4}{e^2}$, with a double root at $x = 2$ and one negative root. For $k = 0$ there is one root ($x = 0$), for $k < 0$ none, and every other positive $k$ gives one or three. So there is 1 such value.

(f) $m x^2 = e^x$ is $x^2 = \dfrac1m e^x$. Three roots need $\dfrac1m < \dfrac{4}{e^2}$, that is $m > \dfrac{e^2}{4} \approx 1.847$. The least integer is $m = 2$. With $m = 1$, $k = 1 > 0.5413$, so there is only one root.