Proving a unique linear transformation is isomorphic I'm very unsure as to whether my proof to the following question make any sense. This is my first time approaching a problem like this, and do to this my thinking about it is prone to error:

My primary concern is with question (b). I will say briefly for (a) that I think that I can show $\mathscr A = \{T_1,T_2,...,T_n\}$ is a basis for $\mathcal L(F,V)$ due to the fact that each $T_i(1)=v_i$ is unique, which means that $Span \ \mathscr A =V$, as every basis vector in $V$ has a unique mapping from $\mathscr A$, so the span of $\mathscr A$ will contain all of $V$. And since $\mathcal L(F,V)$ is the set of all linear transformations from $F \to V$, yet each basis vector of $V$ has one unique linear mapping from $F$ contained in $\mathscr A$,  it can only mean $\mathscr A$ spans $\mathcal L(F,V)$. 
With that out of the way, hopefully I can use this knowledge (or not) to answer (b). 
In order for $_L$ to be an isomorphism, it must be bijective and $dim \ V = dim \ \mathcal L(F,V).$ I will first attempt to prove $_L$ is injective and surjective.
Injectivity
I will prove injectivity by contradiction.
For $_L$ to not be injective, there must exist two vectors in $V$ such that 
$$_L(v_n) =  \ _L(v_m) , \ v_n \neq v_m$$
$$\implies _L(v_n-v_m) = 0 = T_{nm}, \ \ v_n-v_m \in V, T_{nm} \in \mathcal L(F,V)$$
$$ \implies _L(v_n)-_L(v_m) = \sum_{i=1}^n \alpha_i T_i = 0$$
$$\implies \sum_{i=1}^n \beta_i T_i - \sum_{i=1}^n \gamma_i T_i =  \sum_{i=1}^n \alpha_i T_i = 0, \forall \beta, \gamma, \alpha \in F$$
$$\implies \sum_{i=1}^n (\beta_i - \gamma_i) \ T_i = 0$$
However, since $T_i$ is unique, there exists no $T_i$ such that $T_i(1) = 0$ unless $v$ is the $0$ vector of $V$. Hence, the only solution is $\beta_i = \gamma_i$, which implies, $v_n = v_m$ and that this mapping is indeed injective.
Surjectivity
For surjectivity, my answer was unusually short which is making me a bit tentative to believe I truly did prove it.
I will prove it by contradiction.
For the mapping $_L$ not to be surjective, it is required that:
$$\exists \ T_n \in \mathcal L(F,V) : \sum_{i=1}^n \alpha_i \ _L (v_i) \neq T_n$$
As $\sum_{i=1}^n \alpha_i \ _L (v_i) = \mathcal L(F,V).$ However, this is impossible, as $_L(v_i) = T_i$ and so for any $T_n \ \exists \ _L(v_n) $ such that $\sum_{i=1}^n \alpha_i \ _L (v_i) = T_n$. Hence, the mapping is surjective, and thus $_L$ is a bijection. 

All that is left is to show $dim \ V = dim \ \mathcal L(F,V)$. It is understood in the question that $dim \ V = n$. However, also considering $\mathscr A$ is a basis and $T_i(1) = v_i$ this implies for each basis vector in $\mathcal L(F,V)$ (as we've stated that $\mathscr A$ is a basis for $\mathcal L(F,V)$) there is a unique mapping to a basis vector in $V$. Thus, the dimension of $\mathcal L(F,V)$ can be nothing other than $n$. Hence, $_L$ is an isomorphism. 
I'm very inexperienced at proof writing, so please let me know if I made any mistakes, in terms of the process and in terms of little details. 
 A: a) As you implicitly (ought to have) used it, there is a unique linear map $T_v:F\to V$ with $T_v(1)=v$ for any vector $v\in V$, not only for the chosen basis vectors. 
Namely, it is $T_v(\lambda)=\lambda v$.
What you wrote, as it stands, make little sense because at some place you already seem to identify/confuse elements of $\mathcal L(F, V)$ with elements of $V$, and use only the basis vectors instead of generic elements of $V$. 
Instead, you should prove that $T_1,\dots, T_n$ is linearly independent and spans all $\mathcal L(F, V)$. 
 Alternatively, one can prove that actually $v\mapsto T_v$ is an isomorphism, and as such, it automatically takes basis to basis.
(One can observe that this map is just $\iota$, and so in this way, b) implies a).) 
b) To prove a linear map is isomorphism, it suffices to prove any 2 of the 3 conditions you stated, e.g. that it's injective and surjective.
Your proof about injectivity seem to consider only the basis vectors, which is not correct. Nevertheless, using subtraction, we can see that a linear map $f$ is injective iff $f(v)=0\implies v=0$.
So, let $v=\alpha_1v_1+\dots+\alpha_nv_n$. Then $\iota(v)=\dots$, do when it's zero, $\dots$
For surjectivity, let $T\in\mathcal L(F, V) $ arbitrary, then $\dots$ so we found a vector $v\in V$ such that $\iota(v)=T$.
And, by the way, if a) is correctly done, the equality of the dimensions is straightforward, as both spaces have a basis of $n$ elements (meaning that their dimension is $n$).
