the trick in this question is to take the centre of mass (4kN) at the centre of the truss (seeing as its symmetrically loaded). therefore to find reaction at support 1 we take moments about support 2.
Sum of all moments about support 2 = 0 (i will take clockwise rotation as being +)
therefore; 190 * 3 - 2*4000 - 7*450 + 4*(support 1) =0
............support 1 = 10580/4
=2645N
now..to find force in member a, i will do method of sections...taking the section diagonally up through the second member from the left on the top of the truss..and the sum of all the moments about the point directly above member a = 0......therefore taking clockwise rotation as positive we get
0= - 450*2+ A*root3 (simple trig, exact angle triangle)
A= 519.6N (force is towards joint so member is in compression)
sorry its kind of hard to explain without diagrams.