Luckily enough, the angles of the two ropes are the same. Therefore, the tension in each will be the same. This immediately eliminates two of the five answers. Now we just need to calculate what that force is.

We know that together, the vertical components of the tension must equal the weight of the block. Therefore we can write:

$\displaystyle T_1sin(45^{\circ})+T_2sin(45^{\circ}) =mg$

$\displaystyle T_1sin(45^{\circ})+T_2sin(45^{\circ}) = 100N$

Since we know that the two tension forces are equal, we can rewrite:

$\displaystyle 2Tsin(45^{\circ})=100$

Rearranging for T, we get:

$\displaystyle T = \frac{100}{2sin(45^{\circ})} = 70.7 N$

Since there is no friction between the mass and slope, there are only two relevant forces acting on the mass: gravity and tension. Furthermore, since the block is not in motion, we know that these forces are equal to each other. Therefore:

$\displaystyle T = F_g$

Substituting in an expression for the force of gravity, we get:

$\displaystyle T = mgsin(\theta)$

We know all of these values, allowing us to solve for the tension:

$\displaystyle T = (5kg)(10\frac{m}{s^2})sin(50^{\circ})$

$\displaystyle T = 38.3N$

There are three relevant forces acting on the block in this scenario: friction, tension, and gravity. We are given two of these values, so we simply need to develop an expression for the force of gravity in the direction of the slope. Since the block is motionless, we can write:

$\displaystyle F_{net} = 0$

$\displaystyle F_g - T - F_f = 0$

$\displaystyle F_g = T + F_f$

Substituting in an expression for the force of gravity, we get:

$\displaystyle mgsin(\theta) = T+F_f$

Rearrange to solve for the angle:

$\displaystyle \theta = sin^{-1}\left ( \frac{T+F_f}{mg}\right )$

We know all of these values, allowing us to solve:

$\displaystyle \theta = sin^{-1}\left ( \frac{25N+40N}{(10kg)(10\frac{m}{s^2})}\right ) = sin^{-1}(0.65)$

$\displaystyle \theta = 40.5^{\circ}$

There are three relevant forces acting on the block in this scenario: tension, friction, and gravity. We are given tension, so we will need to develop expressions for friction and gravity. Since the block is motionless, we can say:

$\displaystyle F_{net}=0$

$\displaystyle F_g -F_f -T= 0$

Plugging in expressions for the force of gravity and friction, we get:

$\displaystyle mgsin(\theta)-\mu_smgcos(\theta)-T = 0$

Rearranging for the mass, we get:
$\displaystyle m=\frac{T}{g(sin(\theta)-\mu_scos(\theta))}$

We know all of these values, allowing us to solve:

$\displaystyle m = \frac{20N}{(10\frac{m}{s^2})(sin(30^{\circ})-(0.25)cos(30^{\circ}))}$

$\displaystyle m = 7.1 kg$

Since there is no tension, there are only two relevant forces acting on the block: friction and gravity. Since the block is motionless, we can also write:

$\displaystyle F_{net} = 0$

$\displaystyle F_g - F_f = 0$

$\displaystyle F_g = F_f$

Substitute the expressions for these two forces:

$\displaystyle mgsin(\theta) = \mu_s gcos(\theta)$

Canceling out mass and gravitational acceleration, and rearranging for the coefficient of static friction, we get:

$\displaystyle \mu_s = \frac{sin(\theta)}{cos(\theta)}=tan(\theta)$

$\displaystyle \mu_s = 0.36$

There are three relevant forces acting on the block in this situtation: friction, gravity, and tension. We can use Newton's second law to express the system:

$\displaystyle F_{net}=ma$

$\displaystyle F_g -F_f - T = ma$

Substituting expressions in for the forces, we get:

$\displaystyle mgsin(\theta) - \mu_kmgcos(\theta)-T = ma$

Canceling out mass and rearranging to solve for tension, we get:

$\displaystyle T = \frac{gsin(\theta)-\mu_kgcos(\theta)}{a}$

We have values for each variable, allowing us to solve:

$\displaystyle T = \frac{(10\frac{m}{s^2})sin(30^{\circ})-(0.55)(10\frac{m}{s^2})cos(30^{\circ})}{2\frac{m}{s^2}}$

$\displaystyle T = 0.12 N$

In order to find the mass of block 2, we're going to need to calculate a few other things, such as the tension in the rope.

To begin with, we'll need to identify the various forces on our free-body diagram. To do this, we will begin with block 1 and use a rotated coordinate system to simplify things. In such a system, the x-axis will run parallel to the surface of the ramp, while the y-axis will be perpendicular to the ramp's surface, as shown below:

Now we can identify the forces acting on block 1. Along the rotated y-axis, the force of gravity acting on the block is equal to $\displaystyle mg\cos(\theta)$, and the force of the ramp on the block is just the normal force, $\displaystyle N$. Since block 1 is not moving in the y direction, we can set these two forces equal to each other.

$\displaystyle N=m_{1}gcos(\theta )$

Now, considering the forces acting along the rotated x-axis, we have a force pointing downwards equal to $\displaystyle m_{1}g\sin\left ( \theta \right )$. Pointing upwards, we have the tension force $\displaystyle T$ and we also have the frictional force, $\displaystyle F_{k}$.

The formula for calculating the force due to kinetic friction is:

$\displaystyle F_{k}=\mu _{k}N$

Since we have already determined what the normal force is, we can substitute that expression into the above equation to obtain:

$\displaystyle F_{k}=\mu _{k}m_{1}gcos(\theta )$

Now, we can write an expression for the net force acting upon block 1 in the x direction:

$\displaystyle F_{1(net)}=m_{1}a=m_{1}gsin\left ( \theta \right )-T-\mu _{k}m_{1}gcos(\theta )$

Rearrange the above expression to solve for tension.

$\displaystyle T=m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a$

So far, we have only been looking at block 1. Now let's turn our attention to block 2 and see what forces are acting on it. In the downward direction we have the weight of the block due to gravity, which is equal to $\displaystyle m_{2}g$. In the upward direction, as we can see in the diagram, we have the tension of the rope, $\displaystyle T$. We need to write an expression that tells us the net force acting upon block 2.

$\displaystyle F_{2(net)}=m_{2}a=T-m_{2}g$

Since we calculated the expression for tension from the information regarding block 1, we can plug that expression into the above equation in order to obtain:

$\displaystyle m_{2}a=[m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a]-m_{2}g$

Now rearrange to solve for the mass of block 2.

$\displaystyle m_{2}a+m_{2}g=m_{2}(a+g)=m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a$

$\displaystyle m_{2}=\frac{[m_{1}gsin(\theta )-\mu _{k}m_{1}gcos(\theta )-m_{1}a]}{(a+g)}$

Then plugging in values, we can finally calculate block 2's mass:

$\displaystyle m_{2}=\frac{[(12kg*9.8\frac{m}{s^{2}}* sin(30^{o}))-(0.18* 12kg* 9.8\frac{m}{s^{2}}* cos(30^{o}))-(12kg* 1.5\: \frac{m}{s^{2}})]}{(1.5 \frac{m}{s^{2}}+9.8 \frac{m}{s^{2}})}$

$\displaystyle m_{2}=2.28kg$

Since the gravitational force must be cancelled by the tension force, as the ball is experiencing no acceleration, and no other forces are being applied to it:

$\displaystyle F_{grav}=F_{tension}=mg$

$\displaystyle F_{grav}=10kg*9.8 \frac{m}{s^2}=98N= F_{tension}$

The block has three forces on it: the force of tension, the force of gravity, and the force from Bruce. The force of gravity is:

$\displaystyle F_{grav}= mg$

$\displaystyle F_{grav}=10kg*10\frac{m}{s^2}=100N$

The force from Bruce plus the force of tension has to equal gravity (since Bruce's force and tension are up while gravity is down) so the block is in equilibrium. 

$\displaystyle F_{grav}=F_{Bruce}+F_{T}$

$\displaystyle 100N=3N+F_{T}$

$\displaystyle F_{T}=97N$

$\displaystyle F_{total}=F_1+F_2+...$

$\displaystyle ma=F_{rope}+F_{gravity}$

$\displaystyle ma=T+mg$

Plug in values:

$\displaystyle 1500*1=T+1500*-9.8$

Solve for $\displaystyle T$

$\displaystyle T=16.2kN$