To solve the problem, we need to calculate:

1. A vector of magnitude 10 that is parallel to vector $\vec{a} = (-4, 3)$.
2. The unit vector in the opposite direction of $\vec{a}$.

Step 1: Vector of Magnitude 10 Parallel to $\vec{a}$

First, find the magnitude of $\vec{a}$: $\|\vec{a}\| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

To find a vector of magnitude 10 in the same direction, multiply each component by $\frac{10}{5} = 2$: $\vec{v} = 2 \times (-4, 3) = (-8, 6)$

Step 2: Unit Vector in the Opposite Direction of $\vec{a}$

First, find the unit vector in the direction of $\vec{a}$: $\frac{\vec{a}}{\|\vec{a}\|} = \frac{1}{5}(-4, 3) = \left(-\frac{4}{5}, \frac{3}{5}\right)$

Then, reverse the direction: $\text{Opposite Direction Unit Vector} = -\left(-\frac{4}{5}, \frac{3}{5}\right) = \left(\frac{4}{5}, -\frac{3}{5}\right)$

So the answers are:

1. $(-8, 6)$ or $(8, -6)$
2. $\left(\frac{4}{5}, -\frac{3}{5}\right)$