a) Consider $(\mathbb{R}^n, \|.\|_\infty)$ and $(L_b(\mathbb{R}^n, \mathbb{R}),\|.\|)$ which is the space of all linear and bounded functions from $\mathbb{R}^n \to \mathbb{R}$ associated with the operator norm.
$(L_b(\mathbb{R}^n, \mathbb{R}),\|.\|)$ is isomorphic to $\mathbb{R}^n$ if we interpret $A \in (L_b(\mathbb{R}^n, \mathbb{R}),\|.\|)$ as a matrix $(a_1,\dots a_n)$
I want to show that $\|.\|$ is equivalent to $\|.\|_1$ then.
The operator norm is defined as
$$\|Ax\|=\sup\Big\{\frac{\|Ax\|_Y}{\|x\|_X}\:x \in X\setminus\{0\}\Big\}=\sup\{\|Ax\|_Y:x\in X, \|x\|_X=1\}$$
Let $A=(a_1,\dots,a_n)$ and $x=(x_1,\dots,x_n)^T$. Then $Ax=\sum_{i=1}^{n}a_ix_i$.
And $|\sum_{i=1}^{n}a_ix_i|\le\sum_{i=1}^{n}|a_ix_i|\le\max_{i=1,\dots,n}|x_i|\sum_{i=1}^{n}|a_i|=\sum_{i=1}^{n}|a_i|=\|A\|_1$ because $\max_{i=1,\dots,n}|x_i|=1$. Therefore $\|A\|\le\|A\|_1$.
For the other direction $"\|A\|\ge\|A\|_1"$:
Choose $x$ so that $x_i=sgn(a_i), i=1,\dots,n$. Then $\|x\|_\infty=1$ and $|\sum_{i=1}^{n}a_ix_i|=\sum_{i=1}^{n}|a_ix_i|=\sum_{i=1}^{n}|a_i|$. Therefore $\|A\|\ge\|A\|_1$.
Is that correct so far?
b)
To which norm is the operator norm on $L_b(\mathbb{R}^n,\mathbb{R})$ equivalent if we associate $\mathbb{R}^n$ with $\|.\|_2$?
$|\sum_{i=1}^{n}a_ix_i|\le\sum_{i=1}^{n}|a_ix_i|=\sum_{i=1}^{n}\sqrt{(a_ix_i)^2}$ but I don't know how to continue
All norms are equivalent in finite dimensional spaces.