Find the derivative:

\(f'(x)=3x^2-3=3(x-1)(x+1)\).

Critical points: \(x=-1\) and \(x=1\).

Check the sign of the derivative:

for \(x<-1\), \(f'(x)>0\);

for \(-1

for \(x>1\), \(f'(x)>0\).

Therefore, the function increases on \(( -\infty,-1)\) and \((1,+\infty)\), and decreases on \((-1,1)\).

At \(x=-1\), the derivative changes from positive to negative, so this is a maximum point:

\(f(-1)=-1+3+4=6\).

At \(x=1\), the derivative changes from negative to positive, so this is a minimum point:

\(f(1)=1-3+4=2\).

Velocity is the derivative of position with respect to time:

\(v(t)=s'(t)=12+6t\).

According to the condition, \(v(t)=30\), so

\(12+6t=30\).

\(6t=18\).

\(t=3\).

Find the derivative:

\(f'(x)=4x^3-4x=4x(x-1)(x+1)\).

Critical points: \(x=-1,0,1\).

Check the sign of the derivative:

on \(( -\infty,-1)\), \(f'(x)<0\);

on \((-1,0)\), \(f'(x)>0\);

on \((0,1)\), \(f'(x)<0\);

on \((1,+\infty)\), \(f'(x)>0\).

So the function decreases on \(( -\infty,-1)\) and \((0,1)\), and increases on \((-1,0)\) and \((1,+\infty)\).

At \(x=-1\) and \(x=1\), the derivative changes from negative to positive, so these are minima:

\(f(\pm1)=1-2-3=-4\).

At \(x=0\), the derivative changes from positive to negative, so this is a maximum:

\(f(0)=-3\).

A tangent parallel to the line \(y=4x-10\) has slope \(4\).

Find the derivative:

\(f'(x)=4x^3\).

Set the derivative equal to 4:

\(4x^3=4\Rightarrow x^3=1\Rightarrow x=1\).

Find the \(y\)-coordinate of the tangency point:

\(f(1)=1^4+2=3\).

The tangency point is \((1,3)\).

Write the tangent equation:

\(y-3=4(x-1)\).

\(y=4x-1\).

Find the derivative:

\(f'(x)=-3x^2+12x=-3x(x-4)\).

Critical points: \(x=0\) and \(x=4\).

Check the sign of the derivative:

for \(x<0\), \(f'(x)<0\);

for \(00\);

for \(x>4\), \(f'(x)<0\).

Therefore, the function decreases on \(( -\infty,0)\) and \((4,+\infty)\), and increases on \((0,4)\).

At \(x=0\), the derivative changes from negative to positive, so this is a minimum:

\(f(0)=1\).

At \(x=4\), the derivative changes from positive to negative, so this is a maximum:

\(f(4)=-64+96+1=33\).