With questions like this it useful to be familia with the trig ratios of the the angles 30, 45 and 60 degrees, (or their radian equivalents, pi/6, pi/4 and pi/3).
Draw a right-angled triangle with sides adjoining the right-angle 1 and sqrt(3), then Pythagoras will tell you that the length of the hypotenuse is 2.
The angles of the triangle will be 30,60 and 90 degrees.
From that you should be able to read off the sine cosine and tangent (or any of the other three ratios), of 30 or 60 degrees.
In particular cos(60 deg) = 1/2, and sin(60 deg) = sqrt(3)/2.
Now to your actual problem.
The sqrt(3) in the problem should, as they say, 'stick out like a sore thumb'.
If there were a 2 underneath it, it would be the sine of 60 degrees (pi/3 radians).
So,
(1 + sqrt(-3))^5 = (1 + i*sqrt(3))^5 = {2(1/2 + i*sqrt(3)/2)}^5 = 32(cos(60) + i*sin(60))^5.
And now you have to use de'Moivres theorem.
= 32(cos(300) + i*sin(300)).
