A continuous random variable X has the probability density function given by: f(x) = { kx(2-x), 0 <= x <= 2; 0, otherwise. (a) Show that k = 3/4 and compute the exact mean E(X) and variance Var(X). (b) Find the cumulative distribution function F(x) and evaluate the probability P(0.5 <= X <= 1.5).

For $f(x)$ to be a valid probability density function, the integral over its entire domain must equal 1.

$$\int_{-\infty}^{\infty} f(x)\,dx = 1$$

Since $f(x) = kx(2-x)$ for $0 \le x \le 2$ and 0 otherwise, we integrate from 0 to 2:

$$\int_{0}^{2} kx(2-x)\,dx = 1$$ $$k \int_{0}^{2} (2x-x^2)\,dx = 1$$ $$k \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} = 1$$ $$k \left( (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3}) \right) = 1$$ $$k \left( 4 - \frac{8}{3} \right) = 1$$ $$k \left( \frac{12}{3} - \frac{8}{3} \right) = 1$$ $$k \left( \frac{4}{3} \right) = 1 \implies k = \frac{3}{4}$$

Thus, $k = \boxed{\frac{3}{4}}$.

The mean $E(X)$ is given by $\int_{-\infty}^{\infty} x f(x)\,dx$.

$$E(X) = \int_{0}^{2} x \cdot \frac{3}{4} x(2-x)\,dx$$ $$E(X) = \frac{3}{4} \int_{0}^{2} (2x^2 - x^3)\,dx$$ $$E(X) = \frac{3}{4} \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_{0}^{2}$$ $$E(X) = \frac{3}{4} \left( \left( \frac{2(2)^3}{3} - \frac{(2)^4}{4} \right) - (0) \right)$$ $$E(X) = \frac{3}{4} \left( \frac{16}{3} - \frac{16}{4} \right)$$ $$E(X) = \frac{3}{4} \left( \frac{16}{3} - 4 \right)$$ $$E(X) = \frac{3}{4} \left( \frac{16-12}{3} \right) = \frac{3}{4} \left( \frac{4}{3} \right) = 1$$

The exact mean is $E(X) = \boxed{1}$.

First, we compute $E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx$.

$$E(X^2) = \int_{0}^{2} x^2 \cdot \frac{3}{4} x(2-x)\,dx$$ $$E(X^2) = \frac{3}{4} \int_{0}^{2} (2x^3 - x^4)\,dx$$ $$E(X^2) = \frac{3}{4} \left[ \frac{2x^4}{4} - \frac{x^5}{5} \right]_{0}^{2}$$ $$E(X^2) = \frac{3}{4} \left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_{0}^{2}$$ $$E(X^2) = \frac{3}{4} \left( \left( \frac{2^4}{2} - \frac{2^5}{5} \right) - (0) \right)$$ $$E(X^2) = \frac{3}{4} \left( \frac{16}{2} - \frac{32}{5} \right)$$ $$E(X^2) = \frac{3}{4} \left( 8 - \frac{32}{5} \right)$$ $$E(X^2) = \frac{3}{4} \left( \frac{40-32}{5} \right) = \frac{3}{4} \left( \frac{8}{5} \right) = \frac{6}{5}$$

Now, we use the formula $\text{Var}(X) = E(X^2) - (E(X))^2$.

$$\text{Var}(X) = \frac{6}{5} - (1)^2$$ $$\text{Var}(X) = \frac{6}{5} - 1 = \frac{1}{5}$$

The exact variance is $\text{Var}(X) = \boxed{\frac{1}{5}}$.

The CDF $F(x)$ is defined as $F(x) = \int_{-\infty}^{x} f(t)\,dt$.
For $x < 0$:

$$F(x) = \int_{-\infty}^{x} 0\,dt = 0$$

For $0 \le x \le 2$:

$$F(x) = \int_{0}^{x} \frac{3}{4}t(2-t)\,dt$$ $$F(x) = \frac{3}{4} \int_{0}^{x} (2t-t^2)\,dt$$ $$F(x) = \frac{3}{4} \left[ t^2 - \frac{t^3}{3} \right]_{0}^{x}$$ $$F(x) = \frac{3}{4} \left( x^2 - \frac{x^3}{3} \right) = \frac{3x^2}{4} - \frac{x^3}{4}$$

For $x > 2$:

$$F(x) = \int_{0}^{2} \frac{3}{4}t(2-t)\,dt = 1$$

Combining these, the cumulative distribution function is:

$$F(x) = \begin{cases} 0, & x < 0 \\ \frac{3x^2}{4} - \frac{x^3}{4}, & 0 \le x \le 2 \\ 1, & x > 2 \end{cases}$$

We can calculate this probability using the CDF: $P(0.5 \le X \le 1.5) = F(1.5) - F(0.5)$.
Using $F(x) = \frac{3x^2}{4} - \frac{x^3}{4}$:

$$F(1.5) = \frac{3(1.5)^2}{4} - \frac{(1.5)^3}{4} = \frac{3(\frac{9}{4})}{4} - \frac{\frac{27}{8}}{4} = \frac{27}{16} - \frac{27}{32} = \frac{54}{32} - \frac{27}{32} = \frac{27}{32}$$ $$F(0.5) = \frac{3(0.5)^2}{4} - \frac{(0.5)^3}{4} = \frac{3(\frac{1}{4})}{4} - \frac{\frac{1}{8}}{4} = \frac{3}{16} - \frac{1}{32} = \frac{6}{32} - \frac{1}{32} = \frac{5}{32}$$ $$P(0.5 \le X \le 1.5) = F(1.5) - F(0.5) = \frac{27}{32} - \frac{5}{32} = \frac{22}{32} = \frac{11}{16}$$

The probability is $P(0.5 \le X \le 1.5) = \boxed{\frac{11}{16}}$.