For the begining I recall some facts from the theory of general locally convex spaces.
Let $X$ be a linear space with family of seminorms $\{\Vert\cdot\Vert_\lambda:\lambda\in\Lambda\}$. Then $X$ can be made a locally convex topological space. Pre-base of this locally convex topology $\tau$ is given by family of sets $ B_{x_0,\varepsilon,\lambda}=\{x\in X:\Vert x-x_0\Vert_\lambda<\varepsilon\},\quad\text{where}\quad x_0\in X,\;\varepsilon>0,\;\lambda\in L $ In fact topology of each locally convex space can be described by some family of seminorms, and moreover most of definitions in the theory of locally convex can be translated into the language of seminorms.
For example, let $(X,\{\Vert\cdot\Vert_\lambda:\lambda\in L\})$, $(Y,\{\Vert\cdot\Vert_\mu:\mu\in M\})$ and $(Z,\{\Vert\cdot\Vert_\nu:\nu\in N\})$ be locally convex spaces, then bilinear operator $T:X\times Y\to Z$ is continuous iff
$ \forall\nu\in N\quad\exists C>0\quad \exists\lambda_1,\ldots,\lambda_n\in L\quad\exists\mu_1,\ldots,\mu_m\in M\quad $ $ \Vert T(x,y)\Vert_\nu\leq C\max\limits_{i=1,\ldots,n}\Vert x\Vert_{\lambda_i}\max\limits_{i=1,\ldots,m}\Vert y\Vert_{\mu_i} $
Another example: convergence of sequences in $X$ can be described in terms of seminorms $ \lim\limits_{n\to\infty} x_n\underset{\tau}{=}x \quad\Longleftrightarrow\quad \forall\lambda\in\Lambda\quad\lim\limits_{n\to\infty}\Vert x_n-x\Vert_\lambda=0 $
As for the first question, in your situation $ X=C^\infty(\mathbb{R}) $ $ L=\{(K,i)\subset \mathbb{R}:\; K\text{ - is compact},\; i\in\mathbb{N}_0\} $ $ \Vert x\Vert_\lambda=\sup\{|f^{(i)}(t)|:t\in K\},\qquad\lambda=(K,i) $ Now it is remains to recall that convergence in $\sup$ norm is uniform convergence, and the result follows.
As for the second question, in your situation $ X=Y=Z=C^\infty(\mathbb{T}) $ $ L=M=N=\{(K,n)\subset \mathbb{R}:\; K\text{ - is compact},\; n\in\mathbb{N}_0\} $ $ T:C^\infty(\mathbb{T})\times C^\infty(\mathbb{T})\to C^\infty(\mathbb{T}):(x,y)\mapsto xy $ Now take arbitrary $\nu=(K,n)\in N$ and consider $C=2^n$ and $\lambda_i=\mu_i=(K,i)$ with $i=1,\ldots,n$, then $ \begin{align} \Vert T(x,y)\Vert_\nu &=\sup\{|(xy)^{(n)}(t)|:t\in K\}\\ &=\sup\left\{\left|\sum\limits_{i=0}^n {n\choose i}x^{(i)}(t)y^{(n-i)}(t)\right|:t\in K\right\}\\ &\leq\sup\left\{\sum\limits_{i=0}^n {n\choose i}|x^{(i)}(t)||y^{(n-i)}(t)|:t\in K\right\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\sup\{|x^{(i)}(t)||y^{(n-i)}(t)|:t\in K\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\sup\{|x^{(i)}(t)|:t\in K\}\sup\{|y^{(n-i)}(t)|:t\in K\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\Vert x\Vert_{(K,i)}\Vert y\Vert_{(K,n-i)}\\ &\leq 2^n\max\limits_{i=1,\ldots,n}\Vert x\Vert_{\lambda_i}\max\limits_{i=1,\ldots,n}\Vert y\Vert_{\mu_i} \end{align} $ Hence $T$ is continuous.