Solution: The probability density function of $X$ must satisfy \begin{align*} \int\limits_{-\infty } ^{\infty } f(x) = 1 & \implies \int\limits_{0}^{2} \alpha \left( 4x - 2x^2 \right) ~\mathrm{d}x = 1\\ & \implies \alpha \left[ \frac{4x^2}{2} - \frac{2x^3}{3} \right] _{0}^2 = 1 \\ & \implies \alpha \left[ 8 - \frac{16}{3} \right] = 1 \\ & \implies \alpha = \frac{3}{8}. \end{align*} Thus, the function will be \[ f(x) = \begin{cases} \frac{3}{8}\left( 4x - 2x^2 \right) , &\text{ if } 0 \lt x \lt 2 ;\\ 0, &\text{ otherwise} . \end{cases} \]

Now we have \begin{align*} \mathbb{P} \left( \frac{1}{2} \lt X \lt \frac{3}{2} \right) & = \int\limits_{\frac{1}{2}}^{\frac{3}{2}} f(x) \mathrm{d}x \\ & = \int\limits_{\frac{1}{2}}^{\frac{3}{2}} \frac{3}{8} \left( 4x - 2x^2 \right) \mathrm{d}x \\ & = \frac{3}{8} \left[ \frac{4x^2}{2} - \frac{2x^3}{3} \right] _{\frac{1}{2}}^{\frac{3}{2}} \\ & = \frac{3}{8}\left[ \frac{9}{2} - \frac{9}{4} - \frac{1}{2} + \frac{1}{12} \right] \\ & = \frac{11}{16}. \end{align*} Thus, \[ \mathbb{P} \left( \frac{1}{2} \lt X \lt \frac{3}{2} \right) = \frac{11}{16}. \]