The approximate power on the vertical meridian for the Rx +3.00 -1.50 x 102 is?

Vertical meridian is the power in the 90th meridian. First we need to subtract our axis(102 in this problem) from the meridian we are looking for(90 in this problem). This will be $\alpha$ in our oblique meridian formula.

$F_{theta} = ((sin \alpha)^2 \times cyl) + sph$

$F_{theta}$ is the power in the meridian you are trying to find

$\alpha$ is the difference in the axis and the meridian you are trying to find

$F_{cyl}$ is the cylinder of the prescription

$F_{sph}$ is the sphere power of the prescirption

$$ \begin{aligned} F_{theta} &= (sin 12)^2 \times (-1.50) + 3.00\\ &= (0.0432273 \times -1.50) + 3.00\\ &= -0.06484095 + 3.00\\ &= +2.94 \end{aligned} $$

You have a total power of +2.94 D in the $90^{th}$ meridian

If the Rx reads -10.25 -2.50 x 040, what is the approximate power on the horizontal meridian? Horizontal meridian is the power on the $180^{th}$ meridian. In this case you can subtract 40 from 180 or 0 from 40 since it is either 0 or 180. Either way the value will be the same.

$$\begin{align} sin(40)^2 = 0.413176\\ sin(140)^2 = 0.413176 \end{align} $$ $$ \begin{align} F_{theta} &= sin(40)^2 \times (-2.50)) + (-10.25)\\ &= (0.413176 \times (-1.50)) + (-10.25)\\ &= -0.619764 + (-10.25)\\ &= -10.87 D \end{align} $$

You have -10.87 D in the $180^{th}$ meridian