Below follows an alternative analysis which does not rely on Gershgorin's theorem.
We begin the analysis with a sequence of elementary steps designed to reduce the number of variables and simplify the necessary calculations.
In general, we have $$\|A\|_\infty = \|A^T\|_1.$$ If $A$ is symmetric and non-singular, then $A^{-1}$ is also symmetric. It follows that $$ \kappa_\infty(A) = \kappa_1(A).$$ Since our matrix
$$A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$$ is symmetric, we are therefore free to concentrate on, say, the 2-norm and the infinity norm.
We will distinguish between the case of $b=0$ and $b \not = 0$. If $b=0$, then
$$ \|A\|_2 = \|A\|_\infty = \max \{|a|,|c|\}, \|A^{-1}\|_2 = \|A^{-1}\|_\infty = \max \{|a|^{-1},|c|^{-1}\}.$$
It follows, that
$$ \kappa_2(A) = \kappa_\infty(A) = \max\left\{ \frac{|a|}{|c|}, \frac{|c|}{|a|} \right\}$$
In the case of $b = 0$, it is clear that $A$ is well-conditioned precisely when $|a| \approx |c|$ and ill conditioned when $|a| \ll |c|$ or $|c| \ll |a|$.
In the case of $b \not = 0$ we can without loss of generality assume that $b = 1$. If $b \not = 1$, then we simply scale the matrix with $b^{-1}$. The condition numbers are invariant under this scaling because $(b^{-1}A)^{-1} = b A^{-1}$. We can therefore concentrate on the case of $b=1$ where $$A = \begin{bmatrix} a & 1 \\ 1 & c \end{bmatrix}.$$ This matrix is non-singular if and only if $ac \not =1$. In this case, we have
$$A^{-1} = \frac{1}{1 - ac} \begin{bmatrix} c & -1 \\ -1 & a \end{bmatrix}.$$
It follows, that
$$ \|A\|_\infty = \max\{1 + |a|,1+|c|\}, \quad \|A^{-1}\|_\infty = \frac{1}{|1 - ac|}\max\{1 + |a|,1+|c|\},$$
which implies
$$ \kappa_\infty(A) = \frac{1}{|1 - ac|}\max\{(1 + |a|)^2,(1+|c|)^2\}$$
A contour plot of the right hand side could now be obtained. A detailed understanding can be developed by covering the punctured plane $\mathbb{R}^2 - \{(0,0)\}$ with hyperbolas, i.e., curves of the form $$ac = \gamma$$
where $\gamma \not = 0$. On the curve corresponding to a $\gamma \not = 1$ we have
$$ \kappa_\infty(A) \ge \frac{1}{|1 - \gamma|} \left( 1 + \sqrt{|\gamma|}\right)^2$$
with equality achieved when $|a|=|c| = \sqrt{|\gamma|}$. Moreover,
$$ \kappa_\infty(A) \rightarrow \infty, $$
as $$\max\{|a|,|c|\} \rightarrow \infty, \quad ac=\gamma.$$ This covers the analysis of the infinity norm.
Explicit calculation of the 2-norm condition number is tedious, but a short-cut is possible because of the equivalence of norms. In general, we have
$$ \|x\|_\infty \leq \| x\|_2 \leq \sqrt{n} \|x\|_\infty $$
This implies that
$$ \kappa_2(A) \leq n \kappa_\infty(A) \leq n^2 \kappa_2(A)$$
Why? We have
$$ \|Ax\|_2 \leq \sqrt{n} \|Ax\|_\infty \leq \sqrt{n} \|A\|_\infty \|x\|_\infty \leq \sqrt{n} \|A\|_\infty \|x\|_2 $$
This implies that
$$ \|A\|_2 \leq \sqrt{n} \|A\|_\infty$$
Similarly, we have
$$ \|Ax\|_\infty \leq \|Ax\|_2 \leq \|A\|_2 \|x\|_2 \leq \sqrt{n} \|A\|_2 \|x\|_\infty.$$
This implies that
$$ \|A\|_\infty \leq \sqrt{n} \|A\|_2.$$
In our case $n=2$, so
$$ \kappa_2(A) \leq 2 \kappa_\infty(A) \leq 4 \kappa_2(A)$$
or equivalently
$$ \frac{1}{2} \kappa_\infty(A) \leq \kappa_2(A) \leq 2 \kappa_\infty(A). $$
In other words, when it comes to the conditioning of the matrix $A$ there is little to be learned from the 2-norm which cannot be discerned from the infinity norm.
An orthogonal matrix preserves distance (it's an isometry). It means that for any $\mathbf x$ we have $\|Q\mathbf x\|_2=\|\mathbf x\|_2$.
So:
$$\|Q\|_2 \overset{\text{def}}= \sup_{\mathbf x\ne \mathbf 0}\frac{\|Q\mathbf x\|_2}{\|\mathbf x\|_2} = \sup_{\mathbf x\ne \mathbf 0}\frac{\|\mathbf x\|_2}{\|\mathbf x\|_2}=1$$
Since $Q^{-1}$ is orthogonal as well, the result follows.
Best Answer
$\newcommand{\cond}{\operatorname{cond}}$ You have $\cond(A)_2=\frac{\sigma_\max}{\sigma_\min}$ and $\|A\|_2=\sigma_\max$, where the $\sigma$ are the singular values of $A$. This means you need to find a matrix $B$ so that $\|B\|_2=\sigma_\min$. One matrix with that property that comes to mind is $B=u_\min\sigma_\min v_\min^\top$, where $u_k$, $v_k$ are the left and right singular vectors in the SVD $$ A=U\Sigma V^\top=\sum_k u_k\sigma_k v_k^\top. $$ Then $A-B$ has a rank deficiency of one, thus is singular, and $\|B\|_2=\sigma_\min\|u_\min\|\,\|v_\min\|=\sigma_\min$.