for a,
the wind is coming from w45N,
so you just do vector addition and then you get E135S... i think.
I will try to describe: the triangle has the 500km/h going south, and then the 100km/h wind pointing east with an angle of 135deg between the 500km/h (plane) and 100km/h (wind).
Then you just do cosine law, so c = sqrt(a^2+b^2-2(a)(b)cos(C)), sqrt(500^2+100^2-2(500)(100)cos(135)) and you should get 575.7km/h as the resultant.