Alan

9x⁴ - 4y⁴  is the difference of two squares, so can be expressed as:

(3x² + 2y²)(3x² - 2y²)

Turn 6.9% into the fraction 6.9/100 and multiply this by 198:

6.9*198/100 = 13.662

The angle the strings make with the horizontal must be 30°, so the radius of rotation must be

r = 1.59*cos(30°) metres

Resolve forces in vertical and horizontal directions.  These must each be balanced.

Vertical:  T1sin(30°) = T2sin(30°) + mg  ...(1)  

where T1 is tension in upper string (58.2N); T2 that in lower string; m is mass of ball (1.14kg) and g is gravitational acceleration (9.8m/s²)

Horizontal:  T1cos(30°)+T2cos(30°) = mv²/r ...(2) 

where v is speed of ball.

From (1)  T2 = T1 - mg/sin(30°)   ...(3)

                   =  58.2 - 1.14*9.8/sin(30°)

$${\mathtt{T2}} = {\mathtt{58.2}}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}} \Rightarrow {\mathtt{T2}} = {\mathtt{35.856}}$$

T2 ≈ 35.9 Newtons

Use (3) in (2)

T1cos(30°) + T1cos(30°) - mgcos(30°)/sin(30°) = mv²/r 

v² = (r/m)*(2T1cos(30°) - mgcos(30°)/sin(30°))

    = (1.59*cos(30°)/1.14)*(2*58.2*cos(30°) - 1.14*9.8*cos(30°)/sin(30°))

$${\mathtt{v}} = {\sqrt{\left({\frac{{\mathtt{1.59}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{{\mathtt{1.14}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{58.2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}}\right)}} \Rightarrow {\mathtt{v}} = {\mathtt{9.919\: \!048\: \!659\: \!810\: \!567}}$$

v ≈ 9.92 metres/sec

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Let tension in the wire be T, mass be m, velocity be v, length of wire be L, radius be r and gravitational acceleration be g.

Resolve forces horizontally and vertically.

Vertical:  Tcos(θ) = mg  ...(1)

Horizontal:   Tsin(θ) = mv²/r    ...(2)

The radial acceleration is v²/r = Tsin(θ)/m = g*sin(θ)/cos(θ) = g*tan(θ) using (1) and (2)

Plug in the numbers!

The derivation is as follows.  You can plug in the numbers.

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The centrifugal force pushing th box outward is given by m*v²/r where m is the mass of the box, v is the lorry speed and r is the radius of curvature.  This force is balanced by μ*m*g, where μ is the coefficient of static friction and g is the gravitational acceleration.

mv²/r = μmg

v²/r = μg

v² = μgr

v = √(μgr)

You can plug in the numbers.

Identify the forces on each block and use Newton's laws of motion.  Have a look at;

Normal force of surface on lower block = (2.05+4.82)*9.81N,

hence horizontal friction force = 0.257*(2.05+4.82)*9.81N.

Vertical force of upper block on lower block = 2.05*9.81N,

hence horizontal friction force = 0.257*2.05*9.81N.

Newton's 2nd law on lower block:

4.82*a = 106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81

a = (106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81)/4.82

$${\mathtt{a}} = {\frac{\left({\mathtt{106.6}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}\left({\mathtt{2.05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.82}}\right){\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}{\mathtt{2.05}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}{{\mathtt{4.82}}}} \Rightarrow {\mathtt{a}} = {\mathtt{17.450\: \!448\: \!879\: \!668\: \!049\: \!8}}$$

acceleration ≈ 17.45 m/s²

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You write down all the forces on each mass, then resolve them parallel and perpendicular to the surface of the hill.  The perpendicular forces on each mass must balance; the net force down the hill must give rise to an acceleration (the same for both masses since they are rigidly fixed together).  

You don't say which mass is lower so I will assume mass A is lower (this will only affect the sign of the tension);

Mass A:

Perpendicular:   F_NA = m_Ag*cos(θ)                    ...(1)

where F_NA is normal force, m_A is mass of A;  g is gravitational acceleration;  θ is angle to horizontal.

Parallel:           m_A*a = m_Ag*sin(θ) - μ_AF_NA - T  ...(2)

where a is acceleration; μ_A is coefficient of friction for mass A; T is tension in the bar.

Mass B:

Perpendicular:  F_NB = m_Bg*cos(θ)   ...(3)

Parallel:           m_B*a = m_Bg*sin(θ) - μ_BF_NB + T  ...(4)

Use (1) in (2) and (3) in (4)

m_A*a = m_Ag*sin(θ) - μ_Am_Ag*cos(θ) - T     ...(5)

m_B*a = m_Bg*sin(θ) - μ_Bm_Bg*cos(θ)  + T   ...(6)

Add equations (5) and (6):

(m_A + m_B)*a = (m_A + m_B)g*sin(θ) - (μ_Am_A + μ_Bm_B)g*cos(θ)

or

a = g*sin(θ) -  (μ_Am_A + μ_Bm_B)g*cos(θ)/(m_A + m_B)   ...(7)

Substitute (7) back into (5) or (6) and rearrange to get T.  I'll leave you to do this and to plug in the numbers (you should get something close to a ≈ 3.5m/s² and T ≈ -25N)

The normal force is the vertical force the ground exerts on the sled to stop it sinking through the surface!  In this case the sled is partly held up by the force on the rope, so the reaction force of the ground on the sled is

normal force = mass*g - F*sin(θ)  where g is gravitational acceleration (9.8m/s²) and θ is the angle between the rope and the ground. F*sin(θ) is the vertical component of the force applied to the sled through the rope.   If we have θ = 32° then

$${\mathtt{normalforce}} = {\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{32}}^\circ\right)} \Rightarrow {\mathtt{normalforce}} = {\mathtt{218.504\: \!036\: \!788\: \!35}}$$

or normal force ≈ 218.5N