Tension of Each Half (Using Vectors) 
A clothesline is tied between two poles, 8m apart. The line is quite
  taut and has negligible sag. When a wet shirt with a mass of 0.8kg is
  hung at the middle of the line, the midpoint is pulled down 8cm. Find
  the tension in each half of the clothesline.

Most of the examples I have seen for this type of problem are given two angles that can be used to write the vector representation of each of the two tensions, but for this problem, none are given. How does one solve this problem using vectors?
EDIT* inverse trig rules can be used to find the angle
 A: OK, I was about to post my full answer when I had a bit more time, but in fact, you've already linked to it. No matter.
The solution is all about:
a) drawing a free body diagram (FBD) to show the forces acting on one mass of interest - the logical choice here is the wet shirt, which can be treated as a point mass represented by its centre of mass.
b) Coming up with equations to denote the equilibrium of the mass. For most problems of this type, it is easiest to consider two orthogonal (perpendicular) frames - generally horizontal and perpendicular. The mass must be in equilibrium in each frame taken separately. This is the point at which you resolve forces in two orthogonal directions. For convenience, in the vertical frame, denote downward as negative and upward as positive. Similarly, in the horizontal frame, denote leftward as negative and rightward as positive. These are just simple conventions chosen to mirror the usual Cartesian $x-y$ axes.
So, for part a),
Forces acting on mass are:
1) Weight = $-0.8g \ \mathrm{N}$ pointing downward vertically. 
2) Tension from left part of string: $T$  pointing left, but tilted upward at an angle $\theta$ to the horizontal. 
3) Tension from right part of string: $T$  pointing right, but tilted upward at an angle $\theta$ to the horizontal.
Those are the only three forces acting on the mass.
Since the mass is in equilibrium the sum (also known as the resultant) of those three forces is $0$.
Summing vectors at those angles is a bit tough. So the most common procedure is resolution into horizontal and vertical components, as I mentioned above.
In the horizontal frame,
Components of each vector:
1) Weight (horizontal component): $W_H = 0$. Weight is purely vertical, there is no component acting in the horizontal direction, i.e. it is the null vector. In other words, no contribution from weight in the horizontal direction.
2) Tension (left string, horizontal): Consider the right angled triangle formed by the left half of the FBD. The horizontal component of the tension represents one of the perpendicular sides (also known as catheti (singular: cathetus)) of this triangle. The actual tension force represents the hypotenuse of this triangle. You can therefore denote the horizontal component as $T_{l,H} = -T\cos\theta$.
3) Tension (right string, horizontal): Exactly analogous to the above, you can denote the horizontal component here as: $T_{r,H} = +T\cos\theta$
Now note that the sum in the horizontal frame is $0 + (-T\cos\theta) + T\cos\theta = 0$, which doesn't add any information other than show the mass is in equilibrium horizontally. As you would expect.
More rewarding is to consider the vertical frame. Components:
1) Weight: $W_V = -0.8g \ \mathrm{N}$
2) Tension (left string, vertical): $T_{l,V} = +T\sin\theta$ (exactly the same trigonometric approach as above)
3) Tension (right string, vertical): $T_{r,V} = +T\sin\theta$
Now note the sum in the vertical frame: $-0.8g + +T\sin\theta + +T\sin\theta$. This can be set to $0$, since the body is in equilibrium vertically.
So $-0.8g + +T\sin\theta + +T\sin\theta = 0$
$-2T\sin\theta = -0.8g$
$T = \frac{0.8g}{2\sin\theta} = \frac{0.4g}{\sin\theta}$
At this point, you can do as your answer key did. But I prefer to use Pythagoras' theorem. Note that the sides of the triangle (in centimetres) are $8, 400, \sqrt{8^2 + 400^2}$. From that, we can deduce that $\sin\theta = \frac{8}{\sqrt{8^2 + 400^2}} = \frac{8}{\sqrt{160064}}$
Hence $T = \frac{g\sqrt{160064}}{20} \mathrm{N}$
That is an exact answer for the magnitude of the tension force in the string. I believe your answer took $g$ to be $9.81 ms^{-2}$ so we get:
$T \approx 196.24 \mathrm{N}$
That is generally what most physics problems expect you to find. But your answer key expects you to give the answer in Cartesian notation, considering left and right strings separately. That's quite easy.
The vertical component has already been found as: $+0.4g$ in both cases (i.e. half the weight).
The magnitude (absolute value) of the horizontal component can be found by taking $T\cos\theta$, which (making reference to that triangle again) is $T \cdot \frac{400}{\sqrt{160064}}$, and hence this is expressible (after simplification) as $20g$. Of course, the left string has this pointing to the left (negative), and the right string has it to the right (positive).
Putting in the $\vec{i}$ (unit horizontal) and $\vec{j}$ (unit vertical) components as your answer did,
Left string tension = $-20g\vec i + 0.4g \vec j = 196.2 \vec i + 3.924 \vec j$
Right string tension = $20g\vec i + 0.4g \vec j = 196.2 \vec i + 3.924 \vec j$
as required.
