Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix?

Hint: if $A$ and $B$ are square of the same order, $$\det\begin{pmatrix} A & B \\ B & A\end{pmatrix}=\det(A-B)\det(A+B)$$ So setting $$A=\begin{pmatrix}8-\lambda & 0 \\ 0 & 8-\lambda\end{pmatrix}$$

and

$$B=\begin{pmatrix}-8 & 0\\0&-8\end{pmatrix}$$

shows that the determinant of your $4\times 4$ matrix is $(16-\lambda)^2\lambda^2$, and therefore the determinant of your original matrix is $$\boxed{-\lambda^3(16-\lambda)^2}$$

Here's a quick way to do the problem.

We can write this as a block matrix in the following form: $$ B = \pmatrix{ 0&0&0\\ 0&8I&-8I\\ 0&-8I&8I } = 8\pmatrix{ 0&0&0\\ 0&I&-I\\ 0&-I&I } $$ $B$ is a block diagonal matrix; one block is $0$ (corresponding to an eigenvalue of $0$ with corresponding eigenvector $[1,0,0,0,0]^T$), and the other block is $8 \pmatrix{I&-I\\-I&I}$. It suffices to find the eigenvalues of $A = \pmatrix{I&-I\\-I&I}$ and multiply the result by $8$.

In fact, if you are aware of tensor products, you may note that $A = \pmatrix{1&-1\\-1&1} \otimes I$ (in fact, I will be indirectly exploiting this).

Note that the matrix $\pmatrix{1&-1\\-1&1}$ has eigenvalues $0$,$2$ and corresponding eigenvectors $[1,1]^T$ and $[1,-1]^T$.

Now, suppose that $v$ is an eigenvector of $I$ (in other words, let $v$ be any non-zero $2 \times 2$ vector). We find that $$ A \pmatrix{v\\v} = 0 $$ and $$ A \pmatrix{v\\-v} = \pmatrix{2v\\-2v} $$ We may thus establish that $A$ has eigenvectors $0,0,2,2$ As corresponding eigenvectors we may take, for example, $$ v_{0,1} = \pmatrix{1\\0\\1\\0}; v_{0,2} = \pmatrix{0\\1\\0\\1}; v_{2,1} = \pmatrix{1\\0\\-1\\0}; v_{2,2} = \pmatrix{0\\1\\0\\-1} $$

Multiplying the eigenvectors of $A$ by $8$, we note that the matrix $B$ has eigenvectors $0,0,0,16,16$. We may take the eigenvectors $$ \pmatrix{1\\0\\0\\0\\0}; \pmatrix{0\\1\\0\\1\\0}; \pmatrix{0\\0\\1\\0\\1}; \pmatrix{0\\1\\0\\-1\\0}; \pmatrix{0\\0\\1\\0\\-1} $$