The maximum value for sin or cos is 1 and the minimum is -1, therefore the sum of 3sina and 2cosa could not be as high as 13, so I think the question contains an error. The first equation can be solved for a, as follows:
2sina=2-3cosa; 4sin^2a=4-12cosa+9cos^2a; 4-4cos^2a=4-12cosa+9cos^2a; cosa(13cosa-12)=0, and a=90 or 22.62 degrees. a=90 satisfies the equation, but does not satisfy 3sina+2cosa=13, because 3sina+2cosa=3+0=3. a=22.62 doesn't satisfy the first equation.
So 3sina+2cosa=3sina-2cosa=3.
Also:
Square the first equation:
4sin^2a+12sinacosa+9cos^2a=4;
4-4cos^2a+12sinacosa+9-9sin^2a=4;
9-4cos^2a+12sinacosa-9sin^2a=0; 4cos^2-12sinacosa+9sin^2a=9;
(2cosa-3sina)^2=9, so 2cosa-3sina=3, taking square roots. However, -3 is also the square root of 9 and since a=90 is the solution to the first equation, 2cosa-3sina=-3.