Finding a bound on the norm of a matrix first of all I'm sorry but I don't know how to write expressions, so I used images to explain my problem.
I have to find the following parameters: alpha and beta in order that the following bound holds:

where the matrix A is the following one

According to the definition of norm of a matrix, I should find the max eigenvalue of this matrix A.
First of all, I have computed:
Then looking at the solution, I have seen that, the lower 2 × 2 block on the diagonal of this ATA matrix has been called B and the maximum eigenvalue of AT A is:

Question 1: Is the maximum eigenvalue of a matrix always computed in this way? (Cosidering the trace). I usually compute the det(lambda*I-ATA) and then I compute the max eigenvalue
Then in the solution, there was written: Since we are looking for a bound on the norm of A(q), we can write the chain of inequalities

Question 2: Is the bound on the norm of a matrix, always less than the trace of the same matrix?
Question 3: I didn't understand from where the highlighted expressions come up. Could anyone help me?
The final result is the following (but I don't care, I would like to understand the steps):

 A: *

*First, in order to compute the (spectral) norm of $A$, you need to find the maximal eigenvalue of $A^T A$. This is a positive semi-definite matrix so all its eigenvalues are real and non-negative. Your calculation shows that $AA^T$ is block diagonal with zero in the $(1,1)$ entry so the eigenvalues of $AA^T$ are zero and the eigenvalues of $B$ (the lower $2 \times 2$ block). Since $B$ is a $2 \times 2$ matrix, the characteristic polynomial of $B$ is
$$ \det(\lambda I - B) = \lambda^2 - \operatorname{trace}(B) \lambda + \det(B). $$
This is special for $2 \times 2$ matrices. Hence, the eigenvalues of $B$ are
$$ \frac{\operatorname{trace}(B) \pm \sqrt{\operatorname{trace}(B)^2 - 4\det(B)}}{2}$$
and the maximal eigenvalue of $B$ (and hence of $AA^T$) is obtained by taking $+$ in the above formula.

*Since the eigenvalues of $A^TA$ are non-negative, their sum is always greater than or equal to the maximal eigenvalue so we always have $\| A \|^2 \leq \operatorname{trace}(A^T A)$. In your case, $\operatorname{trace}(A^T A) = \operatorname{trace}(B)$.

*The first inequality follows from the fact that $\sin^2 q_2, \cos^2 q_2 \leq 1$. The second inequality follows from
$$ \left( \left( m_2d + m_3|q_3| \right) + \left( \sqrt{2}m_3 \right) \right)^2 = \left( m_2 d + m_3 |q_3| \right)^2 + 2 \left( m_2d + m_3|q_3| \right) \left( \sqrt{2}m_3 \right) + 2m_3^2 $$
and I guess you are assuming that $m_2, m_3, d$ are positive so that you can throw the middle term and replace the first term with $\left( m_2 d + m_3 q_3 \right)^2$ and remain with something smaller.

