= Solution
For the first <matrix>, the <characteristic polynomial> is
$$
\det(tI-A)=(t-3)\big((t-2)(t-1)-6\big)=(t-3)(t-4)(t+1).
$$
Thus the real <eigenvalues> are $4,-1,3$. Corresponding <eigenvectors> are $(3,1,0)^\mathsf T$, $(-2,1,0)^\mathsf T$ and $(0,0,1)^\mathsf T$. They are <linearly independent>; the <determinant> of the <matrix> with these columns is $5$. Therefore
$$
\boxed{X=\begin{pmatrix}3&-2&0\\1&1&0\\0&0&1\end{pmatrix},\qquad X^{-1}AX=\operatorname{diag}(4,-1,3).}
$$
For example, multiplying the first column by $A$ gives $(12,4,0)^\mathsf T$, four times that column; multiplying the second gives $(2,-1,0)^\mathsf T$, its negative. These also check the chosen <eigenvectors> directly.
For the second <matrix>,
$$
\det(tI-A)=t^3-1=(t-1)(t^2+t+1).
$$
The quadratic factor has negative <discriminant> $-3$, so $1$ is the only real <eigenvalue>. Solving $Av=v$ gives $v=(a,a,a)^\mathsf T$, a one-dimensional <eigenspace>. It cannot provide a <basis> of $\mathbb R^3$, so
$$
\boxed{\text{the only real eigenvalue is }1;\quad A\text{ is not diagonalizable over }\mathbb R.}
$$
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