While drawing the free body diagram of the given problem , you will notice that at the moment the jumper is in the air....at that instant, effective acceleration due to gravity would be g*sin600 and along the incline there would be a component of g ,i.e, g*cos600 . After resolving the g into components let us now divide the velocity as such. We will see that along the incline the component of velocity is v*cos300 whereas acting opposite to the effective acceleration due to gravity is a component of v ,i.e., v*sin300 . Successfully resolving these into components we can now calculate the time bettween takeoff and landing as 
Initial velocity in upwards direction being v*sin300 that is, 12.5m/s. Effective g is g*sin600. So, by 
         veffective=ueffective-geffective*t                                     where veffective=0m/s ; ueffective= 12.5m/s; geffective=g*sin600
we get the required time as t=1.47*2=2.94 seconds
Length d of the jump can be calculated by
        s=ueffective*t+0.5*acceleration*t2                here u effective =v*cos300=21.65m/s;t=2.94 s;acceleration=g*cos60=4.9m/s2
so s=d=1948.94m along the inclined plane....if d is measured with respect to ground then that will be dcos300 =1687.78m
maximum distance between the jumper and hill would be given by
                 veffective2  =ueffectrive2-2*geffective*h          where veffective=0m/s;ueffective=12.5m/s;geffective=8.49m/s2   
       so h= 9.20 m
I guess there are no minor mistakes,,, if there were the free body diagram should be clear enough and you may do your calculations yourself. Thank you...