If you have your matrix $A$ given as the following
$$A = \begin{pmatrix} -3 & -4 & -2 \\ 5 & 9 & -5 \\ -3 & 8 & -9\end{pmatrix} $$
then we have
$$ \textrm{a.} \| A\|_{1,1} = \max_{1 \leq j \leq q} \Big\{ \sum_{i=1}^{p} |a_{ij}|\Big\} \textrm{ ( maximum column sum ( modulus))}$$
This is the maximum sum of the absolute values of the columns
$$ \sum_{i=1}^{p} |a_{i1} | = 3 + 5 + 3 = 11 \\ \sum_{i=1}^{p} |a_{i2}| = 4 + 9 + 8 = 21 \\ \sum_{i=1}^{p} |a_{i3}| = 2 + 5+ 9 = 16 $$
which gives us $\|A\|_{1,1} = 21$
$$ \textrm{b.} \| A\|_{\infty,\infty} = \max_{1 \leq i \leq p} \Big\{ \sum_{j=1}^{q} |a_{ij}|\Big\} \textrm{ ( maximum row sum (modulus) )}$$
which is the maximum row sum
$$ \sum_{j=1}^{q} |a_{1j}| = 3 + 4 + 2 = 9 \\ \sum_{j=1}^{q} |a_{2j}| = 5 + 9 + 5 = 19 \\ \sum_{j=1}^{q} |a_{3j}| = 3+8+9 = 20$$
$$ \|A\|_{\infty, \infty} = 20$$
$$ \textrm{c.} \| A \|_{1,\infty} = \max_{i,j} |a_{ij}| $$
which is the largest absolute value of some entry, $a_{22}$ and $a_{33} $ have absolute value $9$
$$ \| A \|_{1,\infty} = 9 $$
$$ \textrm{d.} \| A\|_{2,\infty} = \max_{1 \leq i \leq p} \Big\{ \bigg( \sum_{j=1}^{q} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of rows)}$$
which is the maximum $2$ norm of the rows
$$ \bigg(\sum_{j=1}^{q} |a_{1j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 3^{2} + 4^{2} + 2^{2} } = \sqrt{17} $$
$$ \bigg(\sum_{j=1}^{q} |a_{2j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 5^{2} + 9^{2} + 5^{2} } = \sqrt{131} $$
$$ \bigg(\sum_{j=1}^{q} |a_{3j}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{ 3^{2} + 8^{2} + 9^{2} } = \sqrt{154} $$
we see that $\|A\|_{2,\infty} = \sqrt{154}$
Finally for the last one
$$ \textrm{e.} \| A\|_{1,2} = \max_{j} \Big\{ \bigg( \sum_{i=1}^{p} |a_{ij}|^{2} \bigg)^{\frac{1}{2}} \Big\} \textrm{ ( maximum 2-norm of columns)}$$
is the maximum $2$ norm of the columns
$$ \bigg(\sum_{j=1}^{q} |a_{i1}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{3^{2}+ 5^{2} + 3^{2}} = \sqrt{43} $$
$$ \bigg(\sum_{j=1}^{q} |a_{i2}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{4^{2}+ 9^{2} + 8^{2}} = \sqrt{161} $$
$$ \bigg(\sum_{j=1}^{q} |a_{i3}|^{2} \bigg)^{\frac{1}{2}} = \sqrt{2^{2}+ 5^{2} + 9^{2}} = \sqrt{110} $$
$$ \|A \|_{1,2} = \sqrt{161} $$
Proposition: $\left\lVert UAW\right\lVert=||A||$ if $U$ and $W$ are orthogonal or unitary, for the Frobenius norm and for the operator norm induced by the vector norm $\left\lVert\cdot\right\lVert_2.$
Let's start with the Frobenius norm: Using the trace characterization,
\begin{align*} \left\lVert UAW\right\lVert_F&=\text{tr} (UAWW^TA^TQ^T)=\text{tr}(UAA^TU^T)\\
&=\text{tr}(A^TU^TU^TA)=\text{tr}(AA^T)\\
&=\left\lVert A\right\lVert_F\end{align*}.
Next, for the operator norm induced by the $2$-norm:
\begin{align*}
\left\lVert UAW\right\lVert_2 &=\max_{x\neq 0}\frac{\left\lVert UAWx \right\lVert_2}{\left\lVert x\right\lVert_2}=\max_{x\neq 0}\frac{\sqrt{x^TW^TA^TU^TUAWx}}{\sqrt{x^Tx}}\\
&=\max_{z\neq 0}\frac{\sqrt{zA^TAz}}{\sqrt{z^Tz}}=\max_{z\neq 0}\frac{\left\lVert Az\right\lVert_2}{\left\lVert z\right\lVert_2}=\left\lVert A\right\lVert_2,
\end{align*} where we used the substitution $z=Wx.$
For an example to show that the infinity and $1$-norm do not work, take $U$ to be rotation matrix which rotates counterclockwise by $60$ degrees, $A=\begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},$ and $W=I$. Then, $$UAW=\begin{bmatrix} \frac{1}{2}-\frac{3\sqrt{3}}{2} & 1-2\sqrt{3} \\ \frac{3}{2}+\frac{\sqrt{3}}{2} & 2+\sqrt{3}\end{bmatrix}.$$
Best Answer
We have $x^*y=\langle x, y\rangle$, with $\|x\|_2^2=x^*x=\langle x, x\rangle$, and therefore by Cauchy-Schwartz we obtain $$ |x^*y|\ =\ |\langle x, y\rangle|\ \le\ \|x\|_2\,\|y\|_2$$ So the operator norm of $x^*$ is at most $\|x\|_2$.
On the other hand, taking $y=x$, we get equality, so the norm is precisely $\|x\|_2$.