To prove equivalence of the topologies you can use the fact that both topologies are characterized by their convergent sequences. This is true in any first countable space. Metric spaces are first countable, and the topology induced by a countable family of seminorms is first countable. Here, since you have translation invariance, it is enough to check sequences converging to 0 (although this simplification is not essential). That is, you can show that if $x_1,x_2,\ldots$ is a sequence in your vector space, then $d(x_n,0)\to 0$ as $n\to\infty$ if and only if for all $i$, $p_i(x_n)\to0$ as $n\to\infty$. For the left to right direction, you can use the fact that $\frac{p_i}{1+p_i}\leq 2^id$. For the other direction you can first bound the tail, then work with the remaining finitely many terms.
There is no actual need to consider sequences. You basically want to show that the identity map is a homeomorphism, and this reduces to showing it is continuous at 0 in both directions, which in any case involves the same estimates you would make when working with sequences. (In other words, considering first countability a priori isn't necessary. If you really want to, you could instead work with nets.) Note that Jonas T's answer suggests a complementary approach (that was given around the same time), working directly with neighborhoods of $0$ instead of with sequences.
Hint for exercise 7: Let $\mathcal{Z}$ be the set of complex sequences converging to $0$. Let $b\colon \mathcal{Z} \to [0,1]$ be a bijection. What might be a good choice for the value of $f_n(b(\zeta))$ then?
We choose $f_n(b(\zeta)) = \zeta_n$. If $(\gamma_n)$ is a sequence with $\gamma_n \to \infty$, define $\hat\zeta$ by $\hat\zeta_n = 1/\gamma_n$ if $\gamma_n\neq 0$, and $\hat\zeta_n = 0$ if $\gamma_n = 0$. Then $\gamma_n\cdot f_n(b(\hat\zeta)) = 1$ for all large enough $n$.
Concerning 13 a): No, the distance $d$ is not homogeneous,
$$t\cdot V_\sigma \neq \left\{ f\in C : d(f,0) < t\right\}.$$
If you define $\varphi(x) = \sup \{ \lvert f(x)\rvert : f \in B\}$ for a $\tau$-bounded subset $B\subset C$, what regularity properties of $\varphi$ do you know? And what has Lebesgue to say about the sequence $$\frac{\varphi/n}{1+\varphi/n}\,?$$
$\varphi$ is the supremum of continuous functions, hence it is lower semicontinuous, and therefore measurable. Also, $\varphi$ is non-negative and finite everywhere, since $B$ is $\tau$-bounded, so the sequence $\psi_n = \frac{\varphi/n}{1+\varphi/n} = \frac{\varphi}{n+\varphi}$ of measurable functions converges to $0$ pointwise, and it is evidently dominated by the constant function $1$. By Lebesgue's dominated convergence theorem, or by the monotone convergence theorem, it follows that $\int_0^1 \psi_n(x)\,dx \to 0$. Hence for every $r > 0$, there is an $n_r\in \mathbb{N}\setminus \{0\}$ such that $0 \leqslant \int_0^1 \psi_n(x)\,dx < r$ for all $n \geqslant n_r$. But that implies $n^{-1}\cdot B \subset B_r(0)$ for all $n\geqslant n_r$, so $B$ is shown to be $\sigma$-bounded.
Concerning 13 b): What is your argument for $d(f_n,f) \to 0$ when $f_n \to f$ pointwise?
The dominated convergence theorem. $0 \leqslant \delta_n(x) := \frac{\lvert f_n(x) - f(x)\rvert}{1+\lvert f_n(x) - f(x)\rvert} \leqslant 1$ for all $n$ and $x$, and $\delta_n \to 0$ pointwise. Thus the identity is sequentially continuous (which follows from the fact that it maps $\tau$-bounded sets to $\sigma$-bounded sets).
Why is $\operatorname{id} \colon (C,\tau) \to (C,\sigma)$ not continuous, although it is sequentially continuous?
Because not all $\sigma$-neighbourhoods of $0$ are $\tau$-neighbourhoods of $0$, see next point.
Can you find a neighbourhood $V$ of $0$ in $(C,\sigma)$ such that no $\tau$-neighbourhood of $0$ is contained in $V$?
Any $d$-ball $B_r(0)$ with $0 < r < 1$ will do. Since every $\tau$-neighbourhood of $0$ only constrains the values at finitely many points, every $\tau$-neighbourhood of $0$ contains functions with $d(f,0)$ arbitrarily close to $1$. Take a function that vanishes at the finitely many points controlled by the given $\tau$-neighbourhood, and has sufficiently high triangular spikes in between.
To show that $\tau$ has no countable local basis at $0$, it is not sufficient to expose one local basis that is uncountable. You must show that no countable system can be a neighbourhood basis of $0$.
How does a neighbourhood of $0$ in $\tau$ look, and hence, what can you say about a countable family of neighbourhoods? Which points of $[0,1]$ are controlled by any of these neighbourhoods?
A $\tau$-neighbourhood of $0$ is a set containing $T(\varepsilon; x_1,\dotsc,x_k) = \{ f \in C : \lvert f(x_i)\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant k\}$ for some $\varepsilon > 0$, non-negative integer $k$ and points $x_1,\dotsc,x_k \in [0,1]$. So a countable family $\mathscr{U} = \{U_n : n \in \mathbb{N}\}$ of neighbourhoods ($U_n \supset T(\varepsilon_n;x_1^n,\dotsc x_{k_n}^n)$) only constrains the values at countably many points. For any $x\in [0,1]\setminus \bigcup_n \{x_i^n : 1 \leqslant i \leqslant k_n\}$, none of the $U_n$ is contained in $T(1; x)$, so $\mathscr{U}$ is not a local base. Since $\mathscr{U}$ was an arbitrary countable family of neighbourhoods, it follows that any local base must be uncountable.
Best Answer
Since $f$ is continuous, there exists an open neighborhood $V$ of $0$, such that $$ x\in V \Rightarrow |f(x)|<1. $$ By definition of the topology defined by semi-norms, there exists a finite set $\{p_1,p_2,\ldots,p_n\}\subseteq \mathcal P$, and $\epsilon>0$, such that $$ (\forall i)\ p_i(x)<\epsilon \Rightarrow x\in V. $$ Can you now finish the argument?