1. The direction of the vector [tex]\theta[/tex] relative to the positive x-axis satisfies
[tex]\tan\theta=\dfrac35\implies\theta\approx30^\circ[/tex]
2. A vector with magnitude [tex]\|\mathbf v\|[/tex] and direction [tex]\theta[/tex] has component form
[tex]\mathbf v=\langle\|\mathbf v\|\cos\theta,\|\mathbf v\|\sin\theta\rangle[/tex]
We have [tex]\|\mathbf v\|=5[/tex], but "angle of 60 degrees with the negative x-axis" is a bit ambiguous. I would take it to mean "60 degrees counterclockwise relative to the negative x-axis", so that the direction is [tex]\theta=240^\circ[/tex]. Then
[tex]\mathbf v=\langle5\cos240^\circ,4\sin240^\circ\rangle\approx\langle-2.5,-4.3\rangle[/tex]
But this doesn't match any of the options, so more likely it means the angle is 60 degrees clockwise relative to the negative x-axis, in which case [tex]\theta=120^\circ[/tex] and we'd get
[tex]\mathbf v=\langle5\cos120^\circ,4\sin120^\circ\rangle\approx\langle-2.5,4.3\rangle[/tex]
3. Same as question 2, but now we're using [tex]\mathbf i,\mathbf j[/tex] notation. The vector has magnitude [tex]\|\mathbf v\|=14[/tex] and its direction is [tex]\theta=-30^\circ[/tex]. So the vector is
[tex]\mathbf v=14\cos(-30^\circ)\,\mathbf i+14\sin(-30^\circ)\,\mathbf j=12.1\,\mathbf i-7\,\mathbf j[/tex]
4. The velocity of the plane relative to the air, [tex]\mathbf v_{P/A}[/tex] is 175 mph at 40 degrees above the positive x-axis. The velocity of the air relative to the ground, [tex]\mathbf v_{A/G}[/tex], is 35 mph at 60 degrees above the positive x-axis. We want to know the velocity of the plane relative to the ground, [tex]\mathbf v_{P/G}[/tex].
We use the relationship
[tex]\mathbf v_{P/G}=\mathbf v_{P/A}+\mathbf v_{A/G}[/tex]
Translating the known vectors into component form, we have
[tex]\mathbf v_{P/G}=\langle145\cos40^\circ,145\sin40^\circ\rangle+\langle35\cos60^\circ,35\sin60^\circ\rangle\approx\langle152,143\rangle[/tex]
This vector has magnitude and direction
[tex]\|\mathbf v_{P/G}\|\approx\sqrt{152^2+143^2}\approx208\,\mathrm{mph}[/tex]
[tex]\tan\theta\approx\dfrac{143}{152}\implies\theta\approx43^\circ[/tex]
(or 43 degrees NE)
