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:

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

Acceleration is the derivative of velocity:

\(a(t)=v'(t)=-3+2t\).

According to the condition, \(a(t)=7\), so

\(-3+2t=7\).

\(2t=10\).

\(t=5\).

Find the derivative:

\(f'(x)=-4x^3+16x=-4x(x-2)(x+2)\).

Critical points: \(x=-2,0,2\).

Check the sign of the derivative:

on \(( -\infty,-2)\), \(f'(x)>0\);

on \((-2,0)\), \(f'(x)<0\);

on \((0,2)\), \(f'(x)>0\);

on \((2,+\infty)\), \(f'(x)<0\).

So the function increases on \(( -\infty,-2)\) and \((0,2)\), and decreases on \((-2,0)\) and \((2,+\infty)\).

At \(x=-2\) and \(x=2\), the derivative changes from positive to negative, so these are maxima:

\(f(\pm2)=-16+32-5=11\).

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

\(f(0)=-5\).

The tangent is parallel to the line \(y=2x+8\), so its slope is \(2\).

Find the derivative:

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

Set it equal to 2:

\(-2x+4=2\Rightarrow -2x=-2\Rightarrow x=1\).

Find the tangency point:

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

Write the tangent equation:

\(y-3=2(x-1)\).

\(y=2x+1\).

Find the derivative:

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

Critical points: \(x=-2\) and \(x=2\).

Sign of the derivative:

on \(( -\infty,-2)\), \(f'(x)>0\);

on \((-2,2)\), \(f'(x)<0\);

on \((2,+\infty)\), \(f'(x)>0\).

Therefore, the function increases on \(( -\infty,-2)\) and \((2,+\infty)\), and decreases on \((-2,2)\).

At \(x=-2\) we have a maximum:

\(f(-2)=-8+24+6=22\).

At \(x=2\) we have a minimum:

\(f(2)=8-24+6=-10\).

Velocity is the derivative of position:

\(v(t)=s'(t)=-4+4t\).

According to the condition, \(v(t)=12\), so

\(-4+4t=12\).

\(4t=16\).

\(t=4\).

Find the derivative:

\(f'(x)=8x^3-8x=8x(x-1)(x+1)\).

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

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

At \(x=\pm1\) there are minima:

\(f(\pm1)=2-4+5=3\).

At \(x=0\) there is a maximum:

\(f(0)=5\).

The slope of the given line is \(4\).

Find the derivative of the function:

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

Set the derivative equal to 4:

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

Find the tangency point:

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

Write the tangent equation:

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

\(y=4x+7\).

Find the derivative:

\(f'(x)=3x^2+6x=3x(x+2)\).

Critical points: \(x=-2\) and \(x=0\).

Sign of the derivative:

on \(( -\infty,-2)\), \(f'(x)>0\);

on \((-2,0)\), \(f'(x)<0\);

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

Therefore, the function increases on \(( -\infty,-2)\) and \((0,+\infty)\), and decreases on \((-2,0)\).

At \(x=-2\) there is a maximum:

\(f(-2)=-8+12-4=0\).

At \(x=0\) there is a minimum:

\(f(0)=-4\).

Acceleration is the derivative of velocity:

\(a(t)=v'(t)=-5+2t\).

According to the condition, \(a(t)=11\), so

\(-5+2t=11\).

\(2t=16\).

\(t=8\).

Find the derivative:

\(f'(x)=-12x^3+48x=-12x(x-2)(x+2)\).

Critical points: \(x=-2,0,2\).

Sign of the derivative:

on \(( -\infty,-2)\), \(f'(x)>0\);

on \((-2,0)\), \(f'(x)<0\);

on \((0,2)\), \(f'(x)>0\);

on \((2,+\infty)\), \(f'(x)<0\).

Therefore, the function increases on \(( -\infty,-2)\) and \((0,2)\), and decreases on \((-2,0)\) and \((2,+\infty)\).

At \(x=\pm2\) there are maxima:

\(f(\pm2)=-3\cdot16+24\cdot4-15=-48+96-15=33\).

At \(x=0\) there is a minimum:

\(f(0)=-15\).

The slope of the given line is \(-2\).

Find the derivative of the function:

\(f'(x)=2x+2\).

Set the derivative equal to \(-2\):

\(2x+2=-2\Rightarrow 2x=-4\Rightarrow x=-2\).

Find the tangency point:

\(f(-2)=(-2)^2+2(-2)+5=4-4+5=5\).

Write the tangent equation:

\(y-5=-2(x+2)\).

\(y=-2x+1\).