1. Unique Solution: For which real numbers, a, does the equation a 3^x + 3^-x = 3
have a unique solution?

a=9/4 or a≤0.

Whenever I see the words "unique solution" I start looking for a quadratic to solve, knowing I can throw away everything except the
discriminant, which I set to zero and I'm done. Sure enough, there's a quadratic here, in 3^x, which can be easily solved. As long as a is not zero:

a 3^x + 1/3^x = 3

a 3^2x - 3*3^x + 1 = 0

3^x = (3±sqrt(9-4a))/(2a)

Now, this equation has a unique solution in 3^x when a=9/4.  But what if a=0? Then we have 3^-x=3, which has a unique solution.  So a=9/4 or a=0, and we're done, right? 
Not quite. 3^x must be positive (if x is real) so other values of "a" might exist that give two different roots of the quadratic such that one of the roots must be thrown away, leaving a unique solution. 
This can only happen if one root is positive and the other is not positive. 
There is no value of "a" that can makes a root zero, because of the constant term, so that leaves only the possibility that two non-zero roots have opposite sign, which can only happen if sqrt(9-4a)>3, which can only happen if a<0.

So the final answer is this: a=9/4 or a≤0.

2. Two Solutions: For what values of m does sqrt(x-5)=mx+2 have two solutions?

Before we begin, we should note that m can't be 0, because sqrt(x-5)=2 has
only one solution.

First, square both sides to get a quadratic equation:

sqrt(x-5)=mx+2

x-5=(mx+2)²

x-5=m²x² + 4mx + 4

m²x² + (4m-1)x + 9 = 0    (eq. A1)

Now, find the discriminant to get the range of values of m that gives you two solutions:

(4m-1)²-36(m²) > 0

16m²-8m+1-36m² > 0

20m²+8m-1 < 0

(10m-1)(2m+1) < 0

-1/2 < m < 1/10 and m ≠ 0

But we're not done yet, because we need to test these values of m to make sure they really do yield two solutions.  You see, when we squared both sides of the equation, we might have introduced spurious roots.  Let x₁,x₂ be the two roots of the original equation for a given value of m, with x₁< x₂.  From the quadratic formula we see the roots are:

x₁ = (-4m+1 - sqrt(-20m²-8m+1))/(2m²), and

x₂ = (-4m+1 + sqrt(-20m²-8m+1))/(2m²).

Both sides of the original equation must be positive for both roots. In particular, the larger root, x₂, must be such that mx₂+2
is greater than zero.

mx₂+2 > 0 

m(-4m+1 + sqrt(-20m²-8m+1))/(2m²)+2 > 0 

(1 + sqrt(-20m²-8m+1))/(2m) > 0 

m > 0 

So the final answer is 0 < m < 1/10