Basically just factor the heck out of everything. That includes the leading coefficients. Then nothing is in your way.
f(x) = (5/2) [ x (x  c/5) ]/[x^{2 }+ (a/2)x + (b/2)]
Now you know exactly how everything factors:
 if there's a vertical asymptote at x=1, then there's a factor of (x1) on the bottom.
 If there's a removable discontinuity at x=2, then there's a factor of (x+2) on top and bottom.
So it must be that
f(x) = (5/2) [ x (x+2) ]/[ (x1)(x+2) ]
That's right, there are no coefficients on x because we factored all that out!

Those two versions of f have to match up. So
 c/5 = 2
 x^{2}+(a/2)x+(b/2) = (x1)(x+2) = x^{2}+x2
Now you can easily find a, b, and c.

Making a function continuous just means the function has to be equal to its limit. So just cancel those (x+2)'s and find the limit.
You can see the horizontal asymptote will just be 5/2, since the rational function just averages out x^{2}/x^{2}=1 for large x. (If you want a real calculation, just divide the top and bottom by x^{2} and take the limit as x goes to infinity.