a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{4}{3}x_1'' - \frac{5}{3}x_1'$$
Substitute into the second equation and simplify, we get $$ x_1'' - \frac{5}{2}x_1' + x_1 = 0 $$
which is a second order ODE of $x_1$.
b)
Characteristic equation is $r^2 - \frac{5}{2} r + 1 = (r - \frac{1}{2})(r - 2) = 0$ with roots $r_1 = \frac{1}{2}, r_2 = 2$
General solution for $x_1$ is $x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$
Plug in to $x_2 = \frac{4}{3}x_1' - \frac{5}{3}x_1$ get
$$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
So, $$x_1 = c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$ $$x_2 = -c_1 e^{\frac{1}{2}t} + c_2 e^{2t}$$
Plug in $x_1(0)=-2, x_2(0) = 1$ to get $$-2 = c_1 + c_2 $$ $$1= -c_1 + c_2 $$
Solve the linear system we have
$$c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$$
That is, $$x_1 = -\frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$ $$x_2 = \frac{3}{2} e^{\frac{1}{2}t} -\frac{1}{2} e^{2t}$$
c) See attached picture
Note that as $t \to \infty$, the graph is asymptotic to the line $x_2 = x_1$ in the third quadrant.
