To solve the first equation:

Let's substitute x^2 as y.

So, the equation becomes y^2 - 2y - 3 = 0

Now, we can solve this equation using the quadratic formula:

y = (-(-2) ± sqrt((-2)^2 - 41(-3))) / (2*1)
y = (2 ± sqrt(4 + 12)) / 2
y = (2 ± sqrt(16)) / 2
y = (2 ± 4) / 2

y = (2 + 4) / 2 or y = (2 - 4) / 2

y = 6 / 2 or y = -2 / 2
y = 3 or y = -1

Now, substitute back x^2 for y:

x^2 = 3 or x^2 = -1

This leads to x = √3, x = -√3, x = i and x = -i.

So the solutions to the equation x^4 - 2x^2 - 3 = 0 are x = √3, x = -√3, x = i and x = -i.

To solve the second equation:

8/(x-3) - 10/x = 2

Let's first get rid of the fractions by multiplying through by x(x-3):

8x - 10(x-3) = 2x(x-3)

Expanding:

8x - 10x + 30 = 2x^2 - 6x

Simplifying:

-2x + 30 = 2x^2 - 6x

Rearranging:

2x^2 - 4x - 30 = 0

Dividing by 2:

x^2 - 2x - 15 = 0

Factoring:

(x - 5)(x + 3) = 0

Therefore, the solutions to the second equation 8/(x-3) - 10/x = 2 are x = 5 or x = -3.