By 'being derivable from each other' I assume that you can still use Axioms 2 and 3, but use 'the other' as Axiom 1.
As such, $Q \to (P \to P)$ can be derived from Axioms 2 and 3 together with $P \to (Q \to P)$, i.e. the original Axiom 1:
\begin{array}{lll}
1 & (P \to ((P \to P) \to P) \to ((P \to (P \to P)) \to (P \to P)) &Axiom \ 2\\
2 & P \to ((P \to P) \to P) & Axiom \ 1\\
3 & (P \to (P \to P)) \to (P \to P)& MP \ 1,2\\
4 & P \to (P \to P) & Axiom \ 1\\
5 & P \to P & MP \ 3,4\\
6 & (P \to P) \to (Q \to (P \to P)) & Axiom \ 1\\
7 & Q \to (P \to P) & MP \ 5,6\\
\end{array}
However, $P \to (Q \to P)$ can not be derived from Axioms 2 and 3 together with $Q \to (P \to P)$. Here is why.
Suppose we have a $4$-valued semantics. That is, suppose that all variables take on one of three values $0$, $1$, $2$, or $3$. Now assume that the $\to$ operator works as specified by the following table with the value of the left-operand on the left, and the value of the right-operand at the top:
\begin{array}{c|cccc}
& 0&1&2&3\\
\hline
0&1&1&1&1\\
1&0&1&3&3\\
2&0&1&1&0\\
3&0&1&1&1\\
\end{array}
Thus, for example, $1 \to 2 = 3$
Let's see what happens with this semantics for the $\to$ to the statements of $Q \to (P \to P)$, $((P \to Q) \to P) \to P$, and $P \to (Q \to P)$:
\begin{array}{cc|c|rc|cll|rc}
P&Q&P\to Q & Q \to & (P \to P) & ((P \to Q) & \to P) & \to P & P \to & (Q \to P)\\
0&0&1&\color{red}1&1&1&0&\color{red}1&\color{red}1&1\\
0&1&1&\color{red}1&1&1&0&\color{red}1&\color{red}1&0\\
0&2&1&\color{red}1&1&1&0&\color{red}1&\color{red}1&0\\
0&3&1&\color{red}1&1&1&0&\color{red}1&\color{red}1&0\\
1&0&0&\color{red}1&1&0&1&\color{red}1&\color{red}1&1\\
1&1&1&\color{red}1&1&1&1&\color{red}1&\color{red}1&1\\
1&2&3&\color{red}1&1&3&1&\color{red}1&\color{red}1&1\\
1&3&3&\color{red}1&1&3&1&\color{red}1&\color{red}1&1\\
2&0&0&\color{red}1&1&0&1&\color{red}1&\color{red}1&1\\
2&1&1&\color{red}1&1&1&3&\color{red}1&\color{red}0&3\\
2&2&1&\color{red}1&1&1&3&\color{red}1&\color{red}1&1\\
2&3&0&\color{red}1&1&0&1&\color{red}1&\color{red}1&1\\
3&0&0&\color{red}1&1&0&1&\color{red}1&\color{red}1&1\\
3&1&1&\color{red}1&1&1&3&\color{red}1&\color{red}1&3\\
3&2&1&\color{red}1&1&1&3&\color{red}1&\color{red}0&0\\
3&3&1&\color{red}1&1&1&3&\color{red}1&\color{red}1&1\\
\end{array}
We see that $Q \to (P \to P)$ and $((P \to Q) \to P) \to P$ always evaluate to $1$. As such, we can call them '$1$-tautologies'. But note that $P \to (Q \to P)$ is not a $1$-tautology.
Now, what about the statement $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$? We could do a $64$ row table, but we can reason this out more quickly.
First, since $\varphi \to \psi$ never evaluates to $2$, $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ will never evaluate to $2$.
Second, for the same reason that $\varphi \to \psi$ never evaluates to $2$, we know that $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ cannot evaluate to $3$. This is because the only way for it to evaluate to $3$ would be when $P \to (Q \to R)$ evaluates to $1$, and $(P \to Q) \to (P \to R)$ to $3$, meaning that $P \to Q$ would have to evaluate to $1$, and $P \to R$ to $3$, and that in turn means that $P$ would have to evaluate to $1$ and $R$ to $2$ or $3$. However, with $p$ being $1$, $P \to Q$ can only evaluate to $1$ if $Q$ evaluates to $1$, and with $R$ being $2$ or $3$, that means $Q \to R$ evaluates to $3$, and hence $P \to (Q \to R)$ would evaluates to $3$, rather than $1$. So, $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ cannot evaluate to $3$.
Finally, and again because $\varphi \to \psi$ never evaluates to $2$, the only way for statement $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ to evaluate to $0$ would be when $P \to (Q \to R)$ evaluates to $1$ or $3$, and $(P \to Q) \to (P \to R)$ to $0$, and the latter means that $P \to Q$ would have to evaluate to $1$ or $3$, and $P \to R$ to $0$, and that in turn means that $P$ would have to evaluate to $1$, $2$, or $3$, and $R$ to $0$. This means that $Q$ cannot evaluate to $0$, for otherwise $P \to Q$ would have to evaluate to $0$ as well. But if $Q$ evaluates to $1$, $2$, or $3$, then $Q \to R$ evaluates to $0$, and hence $P \to (Q \to R)$ would evaluate to $0$ as well, so that can't be either. So, there is no way for $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ to evaluate to $0$.
In sum: $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ will always to evaluate to $1$, and is therefore a $1$=-tautology.
OK, so using $Q \to (P \to P)$, $((P \to Q) \to P) \to P$, and $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$ as the axioms, it turns out that they are all $1$-tautologies under this semantics.
Also, note that when applying Modus Ponens to any two $1$-tautologies, you will always get another $1$-tautology, because the only time that $\varphi \to \psi$ and $\varphi$ both evaluate to $1$ is where $\psi$ evaluates to $0$. So, applying Modus Ponens to $1$-tautologies, you can only get more $1$-tautologies.
However, we already saw that $P \to (Q \to P)$ is not a $1$-tautology. Therefore, $P \to (Q \to P)$ cannot be derived from the other three.
