Figured it out!
function IRK(t0, tf, h)
y(1)=1;
t=t0:h:tf;
for i=1:length(t)-1
syms z1 z2
eqn1 = (1-h/4)*z1+h/4*z2==y(i);
eqn2 =(-1)*h/4*z1+(1-5/12*h)*z2==y(i);
zsol = solve([eqn1,eqn2], [z1,z2]);
y(i+1)=y(i)+h/4*(zsol.z1+3*zsol.z2);
end
abs(y(i+1)-exp(1))
So 0.125 as the step size gets closest to the error being $10^{-5}$ . As for the function evaluations I know that each stage of the RK method has two, so 32 all together....But should I include the actual method as two more f evaluations??? I suppose not since the method uses the results from solving the system for each z1...