Cheng Sheng's solution is correct, but here's the typed solution
a)
Let $$P=\begin{pmatrix} 1 & 2 \\ -5 & -1\end{pmatrix}$$
Characteristic polynomial:
$$\chi_P(\lambda) = det\begin{pmatrix} 1-\lambda & 2 \\ -5 & -1-\lambda\end{pmatrix}=\lambda^2 + 9$$
Thus, $$\lambda_1 = 3i, \lambda_2 = -3i$$
Consider $\lambda_1 = 3i$.
$$N(P-3iI) = N\begin{pmatrix} 1-3i & 2 \\ -5 & -1-3i\end{pmatrix} = span\{\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\}$$
Consider $$e^{3it}\begin{pmatrix} -2 \\ 1-3i\end{pmatrix} = \cos(3t) + i\sin(3t)\begin{pmatrix} -2 \\ 1-3i\end{pmatrix}\ = \begin{pmatrix} -2\cos(3t) -2i\sin(3t) \\ \cos(3t) + i\sin(3t) - 3i\cos(3t) + 3\sin(3t)\end{pmatrix} =\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + i \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$
Thus, the general solution of this system is
$$\textbf{x}(t) = c_1\begin{pmatrix} -2\cos(3t) \\ \cos(3t) + 3\sin(3t)\end{pmatrix} + c_2 \begin{pmatrix} -2\sin(3t) \\ \sin(3t) - 3\cos(3t)\end{pmatrix}$$
b)
As $t\to\infty$, solution circulates in clockwise direction in elliptical shapes.
See attached image.
