2.

Multiply by a^x throughout:

Therefore, completing the square;

Rearrange for a^x;

Ok, that's as far as I'm sure for this part. For the other one, I'm not quite getting something.

If z = a^3x = a^-3x;

a^x = z^1/3 = z^-1/3.

Taking a^x = z^1/3;

Expanding, I get:

Maybe it's because I didn't take a^x = z^-1/3, but if I do so, I'll get infinite terms, and I don't know how to deal with that.

3.

Is that it? If so, multiply throughout by (x+1)^1/2;

You can make use of a substitution to help you solve here. Let (x+1)^1/2 = y;

So, 3y -5 = 0,

y = 5/3

and y +1 = 0

y = -1

You can solve those two now.

For the first part, I prefer not answering because I don't quite understand what the question is asking for. I can solve for x, and get x = 2, but not show that it's a real root, sorry