In both cases, you can find these by taking the cross product I L x B, where I is the strength of the current, L is the vector with magnitude the length of the wire, pointing in the direction of the wire, and B is the vector given by the Earth's magnetic field. To find the direction, remember the right hand rule for the cross product.
Going to your specific problems, if we use the usual unit vectors i, j, and k, then
B = .6 * 10^-4 (cos(-75°)i + sin(-75°)j)
part a:
I L = 10*15 i = 150i since we're pointing east, so I L x B = 10*15*.6*10^-4 (i x (cos(-75°)i + sin(-75°)j))
either by the right hand rule or distributing the cross product over addition, (i x (cos(-75°)i + sin(-75°)j)) = sin(-75°)k, so I L x B = 90*10^-4*sin(-75°)k. Since sin(-75°) = -sin(75°) is negative, we end up 9*sin(75°)*10^-3* -k. The direction is therefore the -k direction, which is vertically downward. It's easy to read off the magnitude of the force when we have a single unit vector: it's just 9*sin(75°)*10^-3.
part b:
now I L = 150k, and (k x (cos(-75°)i + sin(-75°)j) = sin(75°)i+cos(75°)j, again by the right hand rule or by distributing. Therefore the force is given by the vector 9*10^-3(sin(75°)i+cos(75°)j). This points 90° above the vector L, so 15° above the horizontal. To find the magnitude, we either use that the vector (x,y) has magnitude sqrt(x^2+y^2), or use that in general the cross product of two vectors a and b has magnitude ||a|| ||b|| sin(theta), where theta is the angle between two vectors. In this case, I L and B are perpendicular, so theta = 90 and sin(theta) = 1, meaning the magnitude is 9*10^-3.
