a) Isolate $x_2$ in equation 1 we get
$$x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$$
Differentiate both sides with respect to $t$ we get
$$x_2' = \frac{1}{2}x_1' - \frac{1}{2}x_1''$$
Substitute into the second equation and simplify, we get $$ x_1'' + 3 x_1' + 2x_1 = 0 $$
which is a second order ODE of $x_1$.
b)
Characteristic equation is $r^2 +3 r + 2 = (r + 2)(r + 1) = 0$ with roots $r_1 = -2, r_2 = -1$
General solution for $x_1$ is $x_1 = c_1 e^{-2t} + c_2 e^{-t}$
Plug in to $x_2 = \frac{1}{2}x_1 - \frac{1}{2}x_1'$ get
$$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
So, $$x_1 = c_1 e^{-2t} + c_2 e^{-t}$$ $$x_2 = \frac{3}{2}c_1 e^{-2t} + c_2 e^{-t}$$
Plug in $x_1(0)=-1, x_2(0) = 2$ to get $$-1 = c_1 + c_2 $$ $$2= \frac{3}{2}c_1 + c_2 $$
Solve the linear system we have
$$c_1 = 6, c_2 = -7$$
That is, $$x_1 = 6 e^{-2t} -7 e^{-t}$$ $$x_2 = 9 e^{-2t} -7 e^{-t}$$
c) See attached image
Note that as $t\to \infty$, the graph approaches the origin in the third quadrant tangent to the line $x_1 = x_2$.
