You may or may not be done, depending on whether you already know that $\{{\neg},{\land},{\lor}\}$ is a functionally complete collection.
If you do know that, you can reason as follows: Suppose we're given some truth function. Express it as an expression that uses $\neg$, $\land$, and $\lor$ (you already know this can be done). Rewrite every $\lor$ in that expression using the proof you already have. The result is an expression for the function that uses only $\neg$ and $\land$.
On the other hand, if you don't know that $\{{\neg},{\land},{\lor}\}$ is complete, then you're not done yet. Your best bet is to know some complete set and show that every function in that set can be made from your selection of operators.
If you don't know any complete sets of connectives, you have to start from scratch and find a proof that for every possible truth table of $n$ inputs can be realized by some expression that uses only $\neg$ and $\land$. (Or perhaps you can show some larger set to be complete, and then use that as a stepping stone to $\{{\neg},{\land}\}$ being complete.
Hint: Consider formulas you can build from $G$ and $\neg$ using only two propositional variables. What can you say about $G(\alpha,\beta,\gamma)$ when $\alpha,\beta,$ and $\gamma$ are each either one of the two propositional variables or their negation?
A stronger hint is hidden below:
Prove by induction that any formula built from $G$ and $\neg$ using only two propositional variables actually depends on only one of the variables.
Best Answer
a)
$\{\vee, \neg\}$ is functionally complete. You only have to show that $\wedge$ (alias $\&$) is definable from $\{\vee, \neg\}$, because then you can express all the connectives of the complete set $\{\vee, \wedge, \neg\}$ (and the functions they induce). Like so: $$ p \wedge q \colon = \neg (\neg p \vee \neg q) $$
b)
$\{\to, \neg\}$ is functionally complete. By a), it suffices to show that $\vee$ is definable from $\{\to, \neg\}$. Thus: $$ p \vee q \colon = (\neg p \to q) $$
c)
$\{\to \}$ is not functionally complete. The proof is by induction on the complexity of propositional formulas in two variables. Let $2 = \{0,1\}$ be the set of truth values. We show that the always-false function $$F_{false} \colon (u,v) \mapsto 0 \colon 2 \times 2 \to 2$$ is not definable. Let $$ \begin{align} F_{\to} &\colon (u, v) \mapsto (\text{$1$ if $u\le v$ else $0$}) \colon 2 \times 2 \to 2 \\ \end{align} $$ $F_{\to}$ is just the truth table for implication "$\to$".
For any propositional formula $A = A(p,q)$ in variables $p, q$, we define the function $f_A \colon 2 \times 2 \to 2$ corresponding to A, inductively:
$$ \begin{align} f_p(u, v) &= u \tag{$pr_1$} \\ f_q(u, v) &= v \tag{$pr_1$} \\ f_{B\to C}(u,v) &= F_{\to}(f_B(u,v), f_C(u,v)) \tag{$Rule_{\to}$} \end{align} $$
Induction on formulas
Atomic formulas are $p$ and $q$. Clearly, neither $f_p$ nor $f_q$ is always false ($0$).
Now suppose $A$ is $B \to C$ with $B, C$ of lesser complexity (height, length). By induction hypothesis, neither $f_B$ nor $f_C$ is always false. Suppose $f_{B \to C}$ is always false: $$f_{B \to C}(u,v) = 0 \;\;\text{for all $u, v$.}$$ By ($Rule_{\to}$), this means: $$F_{\to}(f_B(u,v), f_C(u,v)) = 0 \;\;\text{for all $u, v$.}$$ By the definition of $F_{\to}$, this can only be so when $$ \begin{align} f_B(u,v) &= 1 \;\;\text{for all $u, v$, and} \tag{i} \\ f_C(u,v) &= 0 \;\;\text{for all $u, v$.} \tag{ii} \\ \end{align} $$ But by induction hypothesis, (ii) cannot be.