Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(0≤X≤10)=P(X=0)+P(X=10)
\(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X<0)=P(X=-2)
= 1/4

Given probability distribution function is 
∴ \(F(x)=\left\{\begin{array}{l} 0, \text { if } x<0 \\ \frac{x}{8}, \text { if } 0 \leq x<1 \\ \frac{1}{4}+\frac{x}{8}, \text { if } 1 \leq x<2 \\ \frac{3}{4}+\frac{x}{12}, \text { if } 2 \leq x<3 \\ 1, \text { for } 3 \leq x \end{array}\right.\)
a) i) P(1≤X≤2)
\(\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ \frac { 1 }{ 8 } } dx{ |\frac { 1 }{ 8 } x| }_{ 1 }^{ 2 } } \)
\(=\frac { 1 }{ 8 } (2-1)=\frac { 1 }{ 8 } (1)\)
ii) \(P(x=3)=0\quad if \ \int _{ 3 }^{ 3 }{ f(x)dx=0 } \)
b) X is not a discrete random variable since E is not a step function

We know that,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \)
\(k\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \) 
\(2k\int _{ -\infty }^{ \infty }{ { e }^{ -|x| }dx=1 } \)   (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an function)
\(2k\int _{ 0 }^{ \infty }{ { \left[ \frac { { e }^{ -x } }{ -1 } \right] }_{ 0 }^{ \infty } } =1\)
\(k=\frac { 1 }{ 2 } \)
Mean of the random variable is
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(E(X)=\int _{ -\infty }^{ \infty }{ xk{ e }^{ -|x| }dx } \)    (\(\because { xe }^{ -|x| }\) is an odd function of x)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { xe }^{ -|x| } } \)
= 0
\(E\left( { x }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } f(x)dx\)
\(=\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { ke }^{ -|x| }dx\)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { e }^{ -|x| }dx\)
\(=\int _{ 0 }^{ \infty }{ { x }^{ 2 } } { e }^{ -x }\)  (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an even function)
\(=\Gamma 3\left( \because \Gamma \left( \alpha \right) =\int _{ 0 }^{ \infty }{ { x }^{ \alpha -1 } } { e }^{ -x }dx,\alpha >0;\Gamma n=(n-1)! \right) \)
= 2
\(V(X)=E\left( { x }^{ 2 } \right) -{ \left[ E(X) \right] }^{ 2 }\)
\(=2-{ \left[ 0 \right] }^{ 2 }\)
= 2

From the given data, you first calculate the probability distribution of the random variable. Then using it you calculate mean and variance.
X p(x)
1  F(1) = \(\frac{1}{6}\)
2  F(2)-F(1) = \(\frac{2}{6}\)-\(\frac{1}{6}\) = \(\frac{1}{6}\)
3  F(3)-F(2) = \(\frac{3}{6}\)-\(\frac{2}{6}\) = \(\frac{1}{6}\)
4  F(4)-F(3) = \(\frac{4}{6}\)-\(\frac{3}{6}\) = \(\frac{1}{6}\)
5  F(5)-F(4) = \(\frac{5}{6}\)-\(\frac{4}{6}\) = \(\frac{1}{6}\)
6  F(6)-F(5) = 1-\(\frac{5}{6}\) = \(\frac{1}{6}\)
The probability mass function is

Mean of the random variable X = E(X)\(\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left( 1\times \frac { 1 }{ 6 } \right) +\left( 2\times \frac { 1 }{ 6 } \right) +\left( 3\times \frac { 1 }{ 6 } \right) +\left( 4\times \frac { 1 }{ 6 } \right) +\left( 5\times \frac { 1 }{ 6 } \right) +\left( 6\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } (1+2+3+4+5+6)\)
= 7/2
\(E({ X }^{ 2 })=\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
\(=\left( { 1 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 2 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 3 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 4 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 5 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 6 }^{ 2 }\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 6 }^{ 2 })\)
\(=\frac { 91 }{ 6 } \)
Variance of the Random Variable \(X=V(X)=E\left( { X }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 91 }{ 6 } -{ \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 35 }{ 12 } \)