\(E(X)=\int _{ 0 }^{ 3 }{ x.f(x)dx=\int _{ 0 }^{ 3 }{ x\left( \frac { dx }{ 9 } \right) dx } } \)
\(=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ { x }^{ 2 }dx=\frac { 2 }{ 9 } .{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 } } \)
\(=\frac { 2 }{ 27 } ({ 3 }^{ 3 }-0)=\frac { 2 }{ 27 } (27)=2\)
\(\therefore E(3X+8)=3.E(X)+8\quad [\because E(8)=8]\)
= 3(2) + 8 = 6 + 8
E(3X + 8) = 14

When 2 coins are tossed, sample space S={HH, HT, TH, TT} ⇒ n(s) = 4
∴P(X = 5) = p(getting 2 heads) =\(\frac{1}{4}\)
P(X = 2) = p(getting 1 head) = \(\frac{2}{4}=\frac{1}{2}\)
P(X = 1) = p(getting no head) = \(\frac{1}{4}\)
Hence the probability distribution function is 

\(\therefore E(x)=\sum { { x }_{ i }{ p }_{ i }=1(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 2 } )+5(\frac { 1 }{ 4 } )\)
\(=\frac { 1 }{ 4 } +1+\frac { 5 }{ 4 } =\frac { 1+4+5 }{ 4 } \)
\(=\frac { 10 }{ 4 } =2.50\)
Hence the expected money to win is Rs. 2.50

Given f(x) = 4k(x-1)³,1 ≤ x ≤ 3
Since f(x) is a p.d.f., \(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ 4k({ x-1) }^{ 3 }dx=1\Rightarrow { \left[ \frac { 4k(x-1{ ) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 3 }=1 } \)
\(\Rightarrow k({ 2 }^{ 4 }-{ 0 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
\(p(-2\le x\le 2)=\int _{ -2 }^{ 2 }{ f(x)dx } \)
\(=\int _{ 1 }^{ 2 }{ 4k({ x-1) }^{ 3 }dx } \)
\(=\frac { 4 }{ 16 } \int _{ 1 }^{ 2 }{ { (x-1) }^{ 3 }dx } \)
\([\therefore k=\frac { 1 }{ 16 } ]\)
\(=\frac { 1 }{ 4 } { \left[ \frac { { (x-1) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 16 } ({ 1 }^{ 4 }-0)=\frac { 1 }{ 16 } \)

\(E(X)=\sum { xp(x)=-1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+2\left( \frac { 1 }{ 3 } \right) } \)
\(=\frac { -1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 2 }{ 3 } =\frac { -2+1+4 }{ 6 } \)
\(=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(E({ X }^{ 3 })={ \sum { x } }^{ 2 }p(x)\)
\(=1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 3 } )\)
\(=\frac { 1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 4 }{ 3 } =\frac { 2+1+8 }{ 6 } =\frac { 11 }{ 6 } \)
\(\therefore E{ (2X+3) }^{ 2 }=E(4{ X }^{ 2 }+12X+9)\)
\(=4\left( \frac { 11 }{ 6 } \right) +12\left( \frac { 1 }{ 2 } \right) +9\)
\(=\frac { 22 }{ 3 } +6+9=\frac { 22 }{ 3 } +15\)
\(=\frac { 22+45 }{ 3 } =\frac { 67 }{ 3 } \)
\(\therefore E(2X+3{ ) }^{ 2 }=\frac { 67 }{ 3 } \)

\(E(X)=\sum { xp(x)=-2(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 4 } )+5(\frac { 1 }{ 2 } )\)
\(=\frac { -2 }{ 4 } +\frac { 2 }{ 4 } +\frac { 5 }{ 2 } =\frac { 5 }{ 2 } \)
∴ 4E(X²)-V(2X)=4E(X²)-4.V(X)
=4E(X²)-4[E(X²)-E(X)²]
=4E(X²)-4E(X²)+4[E(X)]²
\(=4{ \left( \frac { 5 }{ 2 } \right) }^{ 2 }[\because E(X)=\frac { 5 }{ 2 } ]\)
\(=4\left( \frac { 25 }{ 4 } \right) =25\)