The cosine of the angle from due West will be the West component (5.66) divided by the total vector size (5.74). 5.66/5.74 is 0.986, and the inverse cosine of this is 9.58 degrees. Adding this to the West direction (270 degrees), we get a final direction of 279.58 degrees.
Given that the original question components were only accurate to one significant figure, your answer of "5.74m/s due 279.58 degrees" would become "6m/s due 280 degrees".
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Problem 2:
Part A:
1000N from the south east towards the north west, plus 1000N from the south west towards the north east.
Vector 1 is 1000N * sin 45 degrees (707N) North plus 1000N * cos 45 degrees (707N) West
Vector 2 is 1000N * sin 45 degrees (707N) North plus 1000N * cos 45 degrees (707N) East
Final vector is 707N + 707N North (total 1414N North) plus zero newtons East/West
= 1414N due North (0 degrees)
The acceleration of the spaceship can be calculated from the equation F=ma
F = 1414N due North
m = 800kg
a = 1414 due North / 800kg = 1.77ms^-2 due North
Given that the original question components were only accurate to one significant figure, your answer of "1.77ms^-2 due North" would become "2ms^-2 due North".
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I note that both of these questions are actually addition of vectors, not subtraction. If you ever need to subtract a vector, simply instead add the equivalent vector in the opposite direction.