1) To solve the equation √(15-x) + √(3-x) = 6, we need to isolate the radical terms on one side and square both sides to eliminate the radicals.

Let's start by rearranging the equation:
√(15-x) = 6 - √(3-x)

Square both sides:
(√(15-x))^2 = (6 - √(3-x))^2
(15-x) = 36 - 12√(3-x) + 3-x

Combine like terms:
15 - x = 39 - 12√(3-x)

Rearrange to isolate the radical term:
12√(3-x) = 24 + x

Square both sides again:
(12√(3-x))^2 = (24 + x)^2
144(3-x) = 576 + 48x + x^2

Expand and simplify:
432 - 144x = 576 + 48x + x^2
0 = x^2 + 192x + 144

Now we have a quadratic equation that we can solve using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
x = (-192 ± √((192)^2 - 41144))/(2*1)
x = (-192 ± √(36864 - 576))/2
x = (-192 ± √(36288))/2
x = (-192 ± 190.5)/2

Now we have two possible solutions for x:
x = (-192 + 190.5)/2 = -1.5/2 ≠ -0.5
x = (-192 - 190.5)/2 = -382.5/2 = -191.25

Therefore, the solution to the equation √(15-x) + √(3-x) = 6 is x = -191.25.

2) To solve the inequality √(x+8) > x+2, we need to isolate the radical term and square both sides to eliminate the radical.

Let's start by isolating the radical term on one side:
√(x+8) > x+2

Square both sides:
(x+8) > (x+2)^2
x+8 > x^2 + 4x + 4

Rearrange to isolate the quadratic terms on one side:
0 > x^2 + 4x + 4 - x - 8
0 > x^2 + 3x - 4

Now we have a quadratic inequality that we can solve by factoring or using the quadratic formula. Factoring, we get:
0 > (x + 4)(x - 1)

Setting each factor to zero to find the critical points:
x + 4 = 0 -> x = -4
x - 1 = 0 -> x = 1

The critical points divide the number line into three intervals: (-∞, -4), (-4, 1), and (1, ∞). Testing a value in each interval will determine if it satisfies the inequality:

1) Choose x = -5 (in interval -∞ to -4):
√(-5+8) > -5+2
√3 > -3 (true)

2) Choose x = 0 (in interval -4 to 1):
√(0+8) > 0+2
√8 > 2 (true)

3) Choose x = 2 (in interval 1 to ∞):
√(2+8) > 2+2
√10 > 4 (true)

Therefore, the solution to the inequality √(x+8) > x+2 is x < -4 or 1 < x < ∞.