I think it would be beneficial to be a bit more explicit in stating the problem. If we have a (not necessarily continuous) linear map $T:(V,|\cdot|_V)\to(W,|\cdot|_W)$ between normed vector spaces over $\mathbb{R}$, then the (operator) norm and conorm of $T$ are defined respectively by
$$\Vert T \Vert = \sup_{|v|_V=1} |Tv|_W\in[0,\infty],\quad \mathfrak{m}(T)=\operatorname{conorm}(T)=\inf_{|v|_V=1}|Tv|_W\in[0,\infty].$$
$\Vert T \Vert\neq\infty$ is equivalent to continuity of $T$. Thus by definition we have
$$\forall v\in V: \mathfrak{m}(T)\, |v|_V\leq |Tv|_W\leq \Vert T \Vert \,|v|_V,$$
and in particular $0\leq \mathfrak{m}(T)\leq \Vert T\Vert\leq \infty$.
Restricting this to finite dimensional $V$ and $W$ (and choosing bases), we have $T:\mathbb{R}^n\to\mathbb{R}^m$. There isn't much to do when either the domain or the codomain is $0$ dimensional, so we may assume $n,m\in\mathbb{Z}_{\geq1}$. We further endow both the domain and the codomain with the Euclidean norms, that is, for $v=(v_1,v_2,...,v_n)\in\mathbb{R}^n$, $|v|=|v|_{\mathbb{R}^n}=\sqrt{\sum_{i=1}^n |v_i|^2}=\sqrt{v_1^2+v_2^2+\cdots+v_n^2}$ (and similarly for $\mathbb{R}^m$). As is customary I'll drop the subscripts of norms going forward (with the caveat that one needs to keep track of the context in which the symbol $|\cdot|$ is used). Again by definition we have
$$\forall v\in \mathbb{R}^n: \mathfrak{m}(T)\, |v|\leq |Tv|\leq \Vert T \Vert \,|v|.$$
Now $\Vert T\Vert\neq\infty$, that is, continuity of $T$, is automatic by finite dimensionality, and so $0\leq \mathfrak{m}(T)\leq \Vert T\Vert< \infty$. Further, the $\sup$ and $\inf$ in the definitions can now be replaced by $\max$ and $\min$, respectively. Note that we could choose to use other norms on the domain and the codomain.
Here are some hints (even to parts the OP doesn't seem to have a problem with):
(a) The last chain of inequalities we wrote above reads $\mathfrak{m}(T)\leq |Tu|\leq \Vert T \Vert$ for $|u|=1$. $\{Tu\,\vert\, |u|=1\}$ is the boundary of the image $TU$ of the unit ball.
It is instructive to draw a picture for a linear map $\mathbb{R}^2\to\mathbb{R}^2$. It's a nice exercise to show that in this case $TU$ is either a point, xor a line segment, xor a filled-in ellipse. Further, the boundary of $TU$ is a circle iff $T$ is a constant multiple of a rotation matrix. Important special cases are when $T$ is represented by a diagonal matrix and when $T$ is represented by a shear.
(b) Similarly the first chain of inequalities we wrote above has the analogous interpretation (even when $T$ is not continuous).
(c) If $T$ is an isomorphism, we have $n=m$. The ($n$-dimensional) volume of the image of the ($n$-dimensional) unit ball under $T$ is $|\det(T)|$ (see e.g. Show that the Area of image = Area of object $\cdot |\det(T)|$? Where $T$ is a linear transformation from $R^2 \rightarrow R^2$), and $T$ is an isomorphism iff $|\det(T)|\neq0$ iff $T$ does not collapse any direction. In particular $TU$ has positive volume (and it's centered at $0$).
(d) For this part we could (and indeed have to, as the next part hints) choose $m\neq n$ to come up with a counterexample, which would guarantee that no $T:\mathbb{R}^n\to\mathbb{R}^m$ is an isomorphism. Say $\mathfrak{m}(T)>0$, so that by part (a) $TU$ has positive ($m$-dimensional) volume. For $m=1$, this means that $TU$ is an interval of the form $[-r,r]$ for some $r\in\mathbb{R}_{>0}$ (indeed by part (a) we have that $r=\Vert T\Vert$). See also the "nice exercise" I mentioned as part of the hint for part (a).
(e) See the hint for part (c).
(f) We know from the preliminary discussion above that $0\leq \mathfrak{m}(T)\leq \Vert T\Vert< \infty$, and the question is asking what $ \mathfrak{m}(T)= \Vert T\Vert$ would imply regarding $T$. The trivial case is when $\mathfrak{m}(T)= \Vert T\Vert=0$; this gives that $T$ is the zero map. It is also straightforward that for $m=1$, $\mathfrak{m}(T)= \Vert T\Vert$ is automatic.
By part (a), note that
$$0<\mathfrak{m}(T) \iff \dim(\operatorname{im}(T))=m,$$
where $\operatorname{im}(T)=\{Tv\in \mathbb{R}^m \,\vert\, v\in \mathbb{R}^n\}$ is the range of $T$. The rank-nullity theorem in tandem with the assumption $ \mathfrak{m}(T)= \Vert T\Vert$ gives that for $n<m$ the trivial case is the only case. Thus for the nontrivial case $ 0<\mathfrak{m}(T)= \Vert T\Vert$ we need to have $T:\mathbb{R}^n\to\mathbb{R}^m$ surjective (and hence $n\geq m$ in particular).
For $n=m=2$, the geometric interpretation of $ \mathfrak{m}(T)= \Vert T\Vert$ is that the largest disk (centered at $0$) that fits inside $TU$ coincides with the smallest disk (centered at $0$) that $TU$ fits inside, so the boundary of $TU$ itself is a circle. Thus $T$ takes the unit circle to some other circle (with possibly zero radius). This heuristic generalizes to higher dimensions with $n=m$.
To deal with the surjective $T$ case with $n>m$ one can add $n-m\in\mathbb{Z}_{>0}$ dimensions to the codomain and replace $T$ with $\widetilde{T}:\mathbb{R}^n\to \mathbb{R}^m\oplus \mathbb{R}^{n-m}, v\mapsto (T(v),0)$. The trick is to use $m$-dimensional volumes in the codomain.
Let me finally note that one could also interpret $\mathfrak{m}(T)= \Vert T\Vert$ in terms of eigenvalues; but it seems the geometric interpretation from part (a) is significantly more convenient in this context than the spectral interpretation.
