Eigen values of $\begin{bmatrix} 0 &5 \\ 5&0 \end{bmatrix}$ are $\pm 5$. Therefore second statement is false.
Since, the rank of matrix $A$ is $2,$ therefore atleast one eigen value would be zero for $n \ge 3$.
For $n= 2,$ It can be proven that $\lambda_1^2 + \lambda_2^2 \le \sum_{i=1}^{n}\sum_{j=1}^{n}A_{ij}^2$.
$\lambda_{1}^{2} + \lambda_{1}^{2} \le 50$
Both $\lambda_{1}$ and $\lambda_{2}$ would be real because $A$ is a real symmetric matrix. Which implies that atleast one eigen value would be in $[-5,5].$
Hence, correct answer is $(B)$
Now, to prove $\lambda_{1}^{2} + \lambda_{2}^{2} \le \sum_{i=1}^{n}\sum_{j=1}^{n}A_{ij}^2$ for $2\times2$ matrix, let us consider the matrix is $\begin{bmatrix} a &c \\ b&d \end{bmatrix}$ and $\lambda$ is the eigen value of this matrix.
${\begin{vmatrix}a-\lambda&c\\ b&d-\lambda\end{vmatrix}} = 0$
$\lambda^{2} - (a+d)\lambda + ad -bc =0$
Let $\lambda_{1}$ and $\lambda_{2}$ are roots of this equation.
$\lambda_{1}^{2} + \lambda_{2}^{2} = (\lambda_{1} + \lambda_2)^{2} - 2 \lambda_{1} \lambda_{2}$
$= (a+d)^{2} -2 (ad-bc)$
$=\sum_{i=1}^{2}\sum_{j=1}^{2}A_{ij}^2 - (b-c)^2$
For real valued matrix,
$\le \sum_{i=1}^{2}\sum_{j=1}^{2}A_{ij}^2 $ (For real symmetric matrix, $b=c$ and $\le$ would be replaced by equal sign)