Given differential equation is \(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}} \)
On squaring both sides, we get 
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2} \)
Here, highest order derivative is \(\frac{d^{2} y}{d x^{2}}\), whose highest power is 2 .
So, order of differential equation is 2 and degree is also 2.

Given, differential equation is 
\(\frac{d^2 y}{d x^2}+\sqrt[3]{\frac{d y}{d x}}+(1+x)=0 \Rightarrow \frac{d^2 y}{d x^2}+(1+x)=-\sqrt[3]{\frac{d y}{d x}}\)
On cubing both sides, we get
\(\left\{\frac{d^2 y}{d x^2}+(1+x)\right\}^3=-\frac{d y}{d x}\)
Hence, the order is 2 and degree is 3. So, the sum is 5.

For independent events,
P(E∩F) = P(E) ⋅ P(F)
\(\Rightarrow 0.35=0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58\)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

Let us denote the numbers on black die by B₁, B₂'.....,B₆ and the numbers on red die by \(R_{1}, R_{2}, \ldots \ldots, R_{6}\) 
Then, we get the following sample space
\(S=\left\{\begin{array}{c} \left(B_{1}, R_{1}\right),\left(B_{1}, R_{2}\right), \ldots \ldots,\left(B_{1}, R_{6}\right),\left(B_{2}, R_{1}\right),\left(B_{2}, R_{2}\right), \ldots \ldots \\ \left(B_{2}, R_{6}\right), \ldots \ldots,\left(B_{6}, R_{1}\right),\left(B_{6}, R_{2}\right), \ldots,\left(B_{6}, R_{6}\right) \end{array}\right)\) 
Clearly, n(S) = 36
Now, let A be the event that sum of number obtained on the die is 8 and B be the event that red die shows a number less than 4.
Then, \(A=\left\{\left(B_{2}, R_{6}\right),\left(B_{6}, R_{2}\right),\left(B_{3}, R_{5}\right),\left(B_{5}, R_{3}\right),\left(B_{4}, R_{4}\right)\right\}\)  
and \(B=\left\{\begin{array}{c} \left(B_{1}, R_{1}\right),\left(B_{1}, R_{2}\right),\left(B_{1}, R_{3}\right),\left(B_{2}, R_{1}\right),\left(B_{2}, R_{2}\right),\left(B_{2}, R_{3}\right) \\ , \ldots . .\left(B_{6}, R_{1}\right),\left(B_{6}, R_{2}\right),\left(B_{6}, R_{3}\right) \end{array}\right\}\) 
\(\Rightarrow A \cap B=\left\{\left(B_{6}, R_{2}\right),\left(B_{5}, R_{3}\right)\right\}\) 
Now, required probability
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}\)

Graph : We have : \(y=\left| x+3 \right| \)
\(=x+3\ ifx\ge -3\)
\(=-x-3\ ifx<-3\)
When \(x\ge -3\)

When x < - 3

Plot the points A (-3, 0), B (-2, 1) and C (-4, 1), D (-5, 2) and portion of the graph is as shown:

Evaluation:
\(\overset { 0 }{ \underset { -6 }{ \int { } } } \left| x+3 \right| dx\)
\(=\overset { -3 }{ \underset { -6 }{ \int { } } } (-x-3)dx+\overset { 0 }{ \underset { -3 }{ \int { } } } (x+3)dx\)
\(=\left[ -\frac { { x }^{ 2 } }{ 2 } -3x \right] _{ -6 }^{ -3 }+\left[ \frac { { x }^{ 2 } }{ 2 } +3x \right] _{ -3 }^{ 0 }\)
\(=\left[ \left( -\frac { 9 }{ 2 } +9 \right) -\left( -\frac { 36 }{ 2 } +18 \right) \right] +\left[ (0+0)-\left( \frac { 9 }{ 2 } -9 \right) \right] \)
\(=\left( \frac { 9 }{ 2 } +0 \right) +\left( 0+\frac { 9 }{ 2 } \right) =9sq.units.\)

The given ellipse is \(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both the axes,

Therefore, area of the ellipse = 4 (Shaded area) = 4 (area OAB)
\(But\ area\ OAB=\overset { 3 }{ \underset { 0 }{ \int { } } } xdy\ [Taking\ horizontal\ strips]\)
\(=\overset { 3 }{ \underset { 0 }{ \int { } } } \frac { 2 }{ 3 } \sqrt { { 9-y }^{ 2 } } dy\)
\([\because \frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { x }^{ 2 } }{ 4 } =1-\frac { { y }^{ 2 } }{ 9 } \Rightarrow x=\frac { 2 }{ 3 } \sqrt { 9-{ y }^{ 2 } } (\because x>0)]\)
\(=\frac { 2 }{ 3 } \left[ \frac { y\sqrt { 9-{ y }^{ 2 } } }{ 2 } +\frac { 9 }{ 2 } { sin }^{ -1 }\frac { y }{ 3 } \right] _{ 0 }^{ 3 }\)
\(=\frac { 2 }{ 3 } \left[ \left[ \frac { 3 }{ 2 } (0)+\frac { 9 }{ 2 } { sin }^{ -1 }(1) \right] -[0-0] \right] \)
\(=\frac { 2 }{ 3 } \left[ \frac { 9 }{ 2 } \left( \frac { \pi }{ 2 } \right) \right] =\frac { 3\pi }{ 2 } \)
\(\therefore From(2),\ area\ of\ the\ ellipse=4\left( \frac { 3\pi }{ 2 } \right) =6\pi sq.units\)

The sample space has 216 outcomes.
\(\text { Now } \quad \mathrm{A}=\left\{\begin{array}{lllll} (1,1,4) & (1,2,4) & \ldots & (1,6,4) & (2,1,4) & (2,2,4) & \ldots (2,6,4) \\ (3,1,4) & (3,2,4) & \ldots &(3,6,4) & (4,1,4) & (4,2,4) & \ldots(4,6,4) \\ (5,1,4) & (5,2,4) & \ldots & (5,6,4) & (6,1,4) & (6,2,4) & \ldots(6,6,4) \end{array}\right\}\)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A ∩ B = {(6,5,4)}.
\(\text { Now }P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216} \)
Then  \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 6 }\)

As shown in the Figure, the line y = 3x + 2 meets x-axis at x\(=\frac{-2}{3}\) and its graph lies below x-axis for \(x \in\left(-1, \frac{-2}{3}\right)\) and above x-axis for \(x \in\left(\frac{-2}{3}, 1\right)\)
The required area = Area of the region ACBA + Area of the region ADEA
\(=\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x\)
\(=\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}\)

The given curves are
y = x² (parabola)
y = x (line)
These intersect at
O(0, 0) and A(1, 1).
 
The area bounded by the curves = Shaded area
 \(=\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) } dx\)
\(=\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } sq.units\)

As favourable cases for sum less than 6 are 10 and favourable for a total of 3 is 2.