The given equations are:

1) |x^2 + 2x + 3| = 3x + 45

2) log3(4x + 15(2x + 27)) = 2log3(4*(2x - 3))

Let's solve equation 1 first:

|x^2 + 2x + 3| = 3x + 45

Since the expression inside the absolute value can be positive or negative depending on the value of x, we will consider both cases:

Case 1: (x^2 + 2x + 3) = 3x + 45
x^2 + 2x + 3 = 3x + 45
x^2 - x - 42 = 0
(x - 7)(x + 6) = 0
Therefore, x = 7 or x = -6

Case 2: -(x^2 + 2x + 3) = 3x + 45
-x^2 - 2x - 3 = 3x + 45
-x^2 - 5x - 48 = 0
(x - 8)(x + 6) = 0
Therefore, x = 8 or x = -6

So, solutions for equation 1 are x = 7, -6, 8.

Now, let's solve equation 2:

log3(4x + 15(2x + 27)) = 2log3(4*(2x - 3))

log3(4x + 30x + 405) = log3((4*(2x - 3))^2)

log3(34x + 405) = log3(16*(2x - 3)^2)

34x + 405 = 16*(2x - 3)^2
34x + 405 = 16(4x^2 - 12x + 9)
34x + 405 = 64x^2 - 192x + 144
64x^2 - 226x + 39 = 0

We can solve this quadratic equation to find the values of x that satisfy equation 2.