Your answer to part (a) is fine.

For part (b), you can do it using the formulae for constant acceleration - have a look at it again and if you still can't do it then let me know what you did and what the 'silly answer' you got was, and I'll try and see what you've done wrong.

Part (c) can also be done with the constant acceleration formulae - in fact I did part (b) by first finding the answer to this part.

Part (d) is harder - here's one way of doing it.

If we call the acceleration down the slope $a$, then we have

$ma=mg\sin \theta -F$

(remember the friction acts in the opposite direction now, as the motion is in the opposite direction), so

$ma=mg\sin \theta -mmg\cos \theta$

$a=g\sin \theta -mg\cos \theta$

$a=g((5/13)-(1/3)(12/13))$

$a=g((5/13)-(4/13))$

$a=g(1/13)=g/13$

Then you can use the appropriate constant acceleration formula to give you $v^2$ and so $v$.

I hope that is of some help. If you're still stuck then please say.

The first way you tried is probably the easiest way of doing this question, but you're just making a small mistake somewhere - the answer should be $650/g$ - here's my working:

$v^2=u^2+2as$

$0={30}^2+2(-9g/13)s$

$18gs/13=900$

$s=900.13/(18g)=100.13/(2g)=(1300/2)/g=650/g$

I hope you can see where you've gone wrong.

In your second method, I don't know what you mean when you write $s=xmg\cos \theta$ - what do you mean by $x$ and by $s$?

Your third method will not work, unfortunately, because in this example, and indeed in any example where there is friction, energy is not conserved. Some of the particle's kinetic energy is used up in overcoming the friction, and will be converted into heat. It is for this reason that the answer to part (d) isn't 30ms ${}^{-1}$, as it would be if there were no friction. (Can you see why this is true?)

Actually, the way I did this question first time round (actually second time, because I messed up the first time :-)) was slightly different (although essentially the same as your first method). I went:

$v=u+\mathrm{at}$

$0=30-(9g/13)t$

$t=130/(3g)$

( $t$ is the time it takes the particle to travel from A to B.)

then

$s=((u+v)/2)t$

$s=15t$

$s=15(130/(3g))$

$s=5.130/g$

$s=650/g$

which (helpfully) gives you the answer to part (c) as well.