Let the velocity of the wind be $\vec{v}_w = (v_{wx}, v_{wy})$ and the velocity of the person be $\vec{v}_p$, both relative to the ground. Let the +x direction be East and +y be North. The apparent wind velocity is $\vec{v}_{app} = \vec{v}_w - \vec{v}_p$.

Case 1: The person runs west at $v_{p1} = 2.5$ m/s.

$\vec{v}_{p1} = (-2.5, 0)$ m/s.

The apparent wind is from the North, so its velocity is South: $\vec{v}_{app1} = (0, -v_{app,y})$.

$\vec{v}_w = \vec{v}_{app1} + \vec{v}_{p1} = (0, -v_{app,y}) + (-2.5, 0) = (-2.5, -v_{app,y})$.

This determines the wind's West-East component: $v_{wx} = -2.5$ m/s.

Case 2: The person's speed doubles to $v_{p2} = 5.0$ m/s, still west.

$\vec{v}_{p2} = (-5.0, 0)$ m/s.

The apparent wind is from the Northwest, so its velocity is Southeast. This means its components are equal and opposite: $\vec{v}_{app2} = (k, -k)$ for some $k > 0$.

$\vec{v}_w = \vec{v}_{app2} + \vec{v}_{p2} = (k, -k) + (-5.0, 0) = (k - 5.0, -k)$.

Comparing the components of $\vec{v}_w$ from both cases:

$$v_{wx} = k - 5.0 = -2.5 \implies k = 2.5 \text{ m/s}$$ $$v_{wy} = -k = -2.5 \text{ m/s}$$

So the wind velocity is $\vec{v}_w = (-2.5, -2.5)$ m/s.

The wind speed is $v_w = \sqrt{v_{wx}^2 + v_{wy}^2} = \sqrt{(-2.5)^2 + (-2.5)^2} = 2.5\sqrt{2}$ m/s. The direction is towards the Southwest, as both components are negative and equal in magnitude.