Closed plus finite dimensional in a TVS If $E$ is a topological vector space (TVS), $F_1$ a closed subspace of $E$, and $F_2$ a finite dimensional subspace of $E$, such that $F_1 \cap F_2=\{0\}$, is $F_1+F_2$ necessarily closed? If yes, are the projection from $F_1+F_2$ onto $F_1$ and $F_2$ respectively, continuous? 

Yes If we suppose that $E$ is a normed space , the sum of a closed subspace $Y$ and a finite dimensional space $F$ is closed, in fact :
suppose that a sequence $(y_n+f_n) \subset Y+F$ converge vers $x\in X$, Let $P$ the (continuous Why? ) projection from $Y+F$ to $F$, the sequence $(y_n+f_n)$ is bounded (because it converge to $x$ so it exist a subsequence $(f_{n_k})$ that converge to $f\in F$ (because F is finite dimensional vector space so every bounded closed set is compact) and so $y_{n_k}$ converge to $y\in Y$ as difference of two convergent sequences, and so $x=y+f\in Y+F$
To complete the proof we need to proof that the  projection from $Y+F$ to $F$ is continuous, so for that we put 
$$
\delta=\min\{d(f,Y) : \|f\|=1 \}
$$
$\delta>0$ because the set $S_F=\{ f\in F ; \|f\|=1\}$ is compact, and the function $d(.,Y)$ is a continuous function so it exist $f'\in S_F$ such that $\delta=d(f',Y)$, then if $\delta=0$ this implies that $f'\in Y\cap F=\{0\}$ but $\|f'\|=1\neq 0$, absurde. 
so 
$$
\|y+f\|\geq \delta\|f\|
$$
so the projection from $Y+F$ to $F$ is of norme $\leq \delta^{-1}$
and the projection from $Y+F$ to $Y$ is of norme $\leq 1+\delta^{-1}$
