Alrighty. For the first I got the solution into the form F(s)= 1/(s^2 + 1) which fits the identity for sin(x). I did this by Taking the Laplace which I gave me:

[s^2*F(s) - f(0) -f'(0)] + F(s) = 0

For the second again I took the Laplace and I got: 6[S^2*F(s)-f(0)-f'(o)] +2[sF(s) - f(0)] +9F(s) = 0

which I simplified into: (6s^2 + 2s +9)F(s) -18 = 0

and further: (6s-2)(s+4)F(s) +17 = 18

and then I had: F(s) = 1/(6s-2)(s+4)

so I use partial fractions: 1= a/(6s-2) + b/(s+4)

I got a=3/13 and b=-1/26

so: F(s) = 3/13(6s-2) + 1/-26(s+4)

Which I think gives: F(t) = 1/26(e^x/3 - e^-4x)

does that look right?