Load factor and how does velocity affect it?

An aircraft initially flying level with a velocity of 130 m/s pulls into:

1. a turn of radius 300m at constant altitude (i.e. a horizontal circle) with constant velocity;

2. a loop (i.e. a vertical circle path) of radius 300m, where the thrust is varied to equal the drag at any point (hence, any change in velocity is entirely due to gravitational effects).

Determine the load factor for (1.) and (2.) when the aircraft is a the 90, 180- and 270-degrees position within the loop.

I have been trying to understand the question, however, I don't entirely understand what they mean. 90, 180 and 270 don't seem like banking angles and specially for part b, does it mean that at 90 degrees it has travelled 300 m vertically and its velocity will decrease based only on the gravity? Also, does the question assume 0 banking angle?

A loop is a vertical circle, so both are not so different from each other.

First, let's check how feasible the assumptions are. The equation for the load factor $$n_z$$ in a horizontal circle is the vector sum of gravity and the required centrifugal force for a circle with speed $$v$$ and radius $$R$$: $$n_z=\sqrt{1+\frac{v^4}{g^2\cdot R^2}}$$ which comes out as 5.83 with the parameters given in your question. This equals a bank angle of 89° and qualifies this circle as a horizontal loop. Note that the limit load factor for aerobatic GA aircraft is $$n_z$$ = 6. There are not many aircraft that can sustain this load factor without altitude loss or full afterburner.

The angular positions mean the stations along the circle: 90° is a quarter circle and 360° is the full circle. Of course, if thrust is sufficient, the load factor will be the same throughout the full circle in case 1 of your question.

Part 2 is not much harder to answer: Here we need to subtract the kinetic energy loss sustained in the climb or regained in the descent phase of the loop. The potential energy gained when climbing 300 m is mass times gravitational acceleration times height, so the speed loss is $$\frac{1}{2}\cdot(v_1^2-v_2^2) = g\cdot h$$ with $$v_1$$ the initial speed of 130 m/s and $$v_2$$ the final speed after the climb by a height $$h$$. Using the values in your question again, the final speed for the first 300 m is 104.597 m/s or about 105 m/s and for the second 300 m it is 71.638 m/s. So the answer for the speed is 105 m/s at 90° and 270° and 71.64 m/s at 180°.

Now use those speeds to calculate the load factor at those points. Hint: At the start of the loop it is $$n_z$$ = 6.74.

b) means straight up into a vertical loop with a 300m radius, with the load factor calculated at the 9 o'clock position as velocity is decreasing and G increasing on the up side, 12 o'clock position as velocity and G reach a minimum at the top inverted, and 3 o'clock position as velocity and G is increasing on the down side, all due to gravity alone. Bank angle is not a factor and can be considered "level" flight from that perspective.

a) means the load factor in a 300 m radius level turn, at whatever bank angle is required to achieve that radius at a constant 130 m/s.

The question should end with "loop and turn" instead of just loop and I would assume the 90, 180, 270 business also applies to the equivalent locations in the level turn in the case of a) (and I would presume to be the same number at each location in the level turn).

The question is pretty poorly done, but that's how I would interpret it to make do with what is presented. Between the confusion over loop and turn and the typos like "id" for "is", the exam writer needs some remedial training. So if you get the answer wrong, and it's important, I would protest it if that is feasible.

I don't entirely understand what they mean. 90, 180 and 270 don't seem like banking angles

They're not. They're the position within the loop. Where the question reads:

Determine the load factor for (1.) and (2.) when the aircraft is a the 90, 180- and 270-degrees position within the loop.

I would suggest parsing it as: