For the first problem:
The ratio g(0)/g(-2) = 1/2
This suggests that we look for g of the form g(x) = A 2^{-x/b} for some value of A and b.
We know there should be a minus sign in the exponent because g(0) is less than g(-2). Of course A cancels out in the ratio g(0)/g(-2). Thus that ratio is just 2^{0}/2^{2/b} which must evaluate to 1/2. By inspection we see that
this implies that b = 2. So g(x) = A 2^{-x/2} for some value of A. However,
g(0) = A 2^{-0/2} = 6. So A = 6. Finally
g(x) = 6 2^{-x/2}
For the second problem:
h(.5) / h(.25) = 1/4 = 1/2^{2}
As above this suggests that we look for h of the form h(x) = B 2^{-x/d} for some value of B and d.
Proceeding as above h(.5)/h(.25) = 2^{-.5/d}/2^{-.25/d} = 2^{-.25/d} = 1/2^{2} , by inspection d = .125
So h(x) = B 2^{-8 x} and h(.5) = B 2^{-4} which must equal 4. Thus B = 64.
Finally h(x) = 64 2^{-8 x}
Along the way I used the properties of exponential forms and 1/.125 = 8