Given F(5) = (k+2)/(4).

Step 1: Determine the value of $k$.
For a cumulative distribution function (CDF) of a discrete random variable, the probability of the last value must be 1.
Given $F(5) = \frac{k+2}{4}$.

$$F(5) = 1$$ $$\frac{k+2}{4} = 1$$ $$k+2 = 4$$ $$k = 4-2$$ $$k = 2$$

This shows that $k=2$.

Step 2: Find $P(1 \le X \le 4)$ in terms of $k$.
For a discrete random variable, $P(a \le X \le b)$ is the sum of probabilities for $X$ values from $a$ to $b$.
The given table provides the cumulative distribution function $F(x) = P(X \le x)$.
Therefore, $P(1 \le X \le 4)$ is equivalent to $P(X \le 4)$, since 1 is the smallest value $X$ can take.

$$P(1 \le X \le 4) = F(4)$$

From the table, $F(4) = \frac{k+5}{10}$.

$$P(1 \le X \le 4) = \frac{k+5}{10}$$

Using the value of $k=2$:

$$P(1 \le X \le 4) = \frac{2+5}{10} = \frac{7}{10}$$

Step 3: Construct the probability mass function (PMF) using $k=2$.
The values of $F(x)$ with $k=2$ are:
$F(1) = \frac{3}{10}$
$F(2) = \frac{2(2)+3}{20} = \frac{4+3}{20} = \frac{7}{20}$
$F(4) = \frac{2+5}{10} = \frac{7}{10}$
$F(5) = \frac{2+2}{4} = \frac{4}{4} = 1$

Now, calculate the individual probabilities $P(X=x)$:
$P(X=1) = F(1) = \frac{3}{10}$
$P(X=2) = F(2) - F(1) = \frac{7}{20} - \frac{3}{10} = \frac{7}{20} - \frac{6}{20} = \frac{1}{20}$
$P(X=4) = F(4) - F(2) = \frac{7}{10} - \frac{7}{20} = \frac{14}{20} - \frac{7}{20} = \frac{7}{20}$
$P(X=5) = F(5) - F(4) = 1 - \frac{7}{10} = \frac{3}{10}$

The PMF is:

$x$	$P(X=x)$
1	$\frac{3}{10}$
2	$\frac{1}{20}$
4	$\frac{7}{20}$
5	$\frac{3}{10}$

Step 4: Calculate $E(X)$.

$$E(X) = \sum x \cdot P(X=x)$$ $$E(X) = 1 \cdot \frac{3}{10} + 2 \cdot \frac{1}{20} + 4 \cdot \frac{7}{20} + 5 \cdot \frac{3}{10}$$ $$E(X) = \frac{3}{10} + \frac{2}{20} + \frac{28}{20} + \frac{15}{10}$$ $$E(X) = \frac{6}{20} + \frac{2}{20} + \frac{28}{20} + \frac{30}{20}$$ $$E(X) = \frac{6+2+28+30}{20} = \frac{66}{20} = \frac{33}{10}$$ $$E(X) = 3.3$$

Step 5: Calculate $Var(X)$.
First, calculate $E(X^2)$:

$$E(X^2) = \sum x^2 \cdot P(X=x)$$ $$$$