(a) If $f\::\:R\:\rightarrow \:T$ is an epimorphism, and $I\subset R$ is an ideal, then $f\left(I\right)$ is an ideal in T
(b) If I take an ideal $J\supset I$, and $\pi :\:R\:\rightarrow \:R/I$ is an epimorphism, then $\pi \left(J\right)$ is also an ideal
(c) The function $\pi$ is a bijection which retains the inclusion relation
(d) $\left\{J\supset I\::\:J\:\in Max\:R\right\}\ni L\:\rightarrow \:\pi \left(L\right)\in \:\left\{M\:\subset R/I\::\:M\:\in \:Max\:R/I\right\}$ is also a bijection which retains the inclusion relation (the same goes for Spec R and Spec R/I)
(e) $I \in Max\:R\:\Leftrightarrow \:\pi \left(I\right)=\left(0\right)\in Max R/I \Leftrightarrow R/I\text{ is a field}$
My approach to that is:
(a) and (b) essentially describe the same problem because we are dealing with an epimorphism and this can be proved by just writing out the properties of ideals and using the homomorphism of f and $\pi$.
What I don't understand though is (c), why should $\pi$ be a bijection if we defined it as an epimorphism which is a surjection, but no injection? Also what does it mean to "retain the inclusion relation" (that's translated from my mother tongue so maybe it's different in English)
I'd really like to see how we should prove (d), I am fairly confident that we should use the homomorphism properties of $\pi$ but how?
I kind of understand the proof of (e), because you use the fact that if I is a maximal ideal then R/I is a field, but why is $\pi \left(I\right)=\left(0\right)\in Max R/I$? That's something I don't quite see
For the purposes of part (a), we need to be clear that we're referring to a surjective ring homomorphism. The term "epimorphism" generally does not mean "surjective ring homomorphism" - this is not standard terminology. But it's the OP's definition, so I'll go with it.
Just to make things clearer, I'm going to write $Im_\pi(S)$ for the image of $S$ under $\pi$ (what you're calling $\pi(S)$).
Part (c) says that $Im_\pi$ is a bijection between ideals of $R / I$ and ideals of $R$ which contain $I$. It does not say that $\pi$ itself is a bijection.
"Maintaining the inclusion relation" means that $Im_\pi(J) \subseteq Im_\pi(K)$ iff $J \subseteq K$ for all ideals $J, K$ such that $I \subseteq J$ and $I \subseteq K$.
Regarding (d), we prove it as follows:
Suppose that $L$ is a maximal ideal of $R$ and $I \subseteq L$. Then $Im_\pi(L)$ is an ideal of $R / I$. Suppose we have an ideal $Z \subseteq R / I$ such that $Im_\pi(L) \subseteq Z \subsetneq R/I$. Then we can write $Z = Im_\pi(Y)$ for some ideal $Y$ such that $I \subseteq Y$. Then we have $Im_\pi(L) \subseteq Im_\pi(Y) \subsetneq Im_\pi(R)$, and since $Im_\pi$ preserves the $\subseteq$ relation in both directions, we have $L \subseteq Y \subsetneq R$. Then $Y = L$. Then $Z = Im_\pi(L)$. Thus, we see that $Im_\pi(L)$ is a maximal ideal.
Now suppose that $L$ is an ideal of $R$ such that $I \subseteq L$, and suppose that $Im_\pi(L)$ is a maximal ideal. Suppose we have $L \subseteq K \subsetneq R$ for some ideal $K$. Then we have $Im_\pi(L) \subseteq Im_\pi(K) \subsetneq Im_\pi(R) = R / I$, since $Im_\pi$ preserves the $\subseteq$ relation and is injective. Then $Im_\pi(K) = Im_\pi(L)$, since $Im_\pi(L)$ is a maximal ideal. Then $K = L$, since $Im_\pi$ is injective.
Thus, we see that $Im_\pi$ restricts to a bijection $\{L$ a maximal ideal of $R$: $I \subseteq L\} \to \{K$ a maximal ideal of $R / I\}$. Since $Im_\pi$ preserves $\subseteq$ in both directions, so does the restriction.
For proving (d) with respect to prime ideals:
Suppose $I \subseteq L$, $L$ is a prime ideal. Now suppose we have $xy \in Im_\pi(L)$. Then we write $x = \pi(a)$, $y = \pi(b)$. Then we have $\pi(ab) \in Im_\pi(L)$. Then we can write $ab = \ell + i$ where $\ell \in L$ and $i \in I$. But we see that since $I \subseteq L$, we have $\ell + i \in L$. Then $ab \in L$. Then $a \in L$ or $b \in L$. Then $\pi(a) \in Im_\pi(L)$ or $\pi(b) \in Im_\pi(L)$. Then $Im_\pi(L)$ is prime.
Now suppose $I \subseteq L$, $L$ is an ideal, and $Im_\pi(L)$ is prime. Consider $a, b$ such that $ab \in L$. Then $\pi(a) \pi(b) \in Im_\pi(L)$. Then $\pi(a) \in Im_\pi(L)$ or $\pi(b) \in Im_\pi(L)$. WLOG, take $\pi(a) \in Im_\pi(L)$. Then we can write $\pi(a) = \pi(\ell)$ for some $\ell \in L$. Then $a = \ell + i$ for some $i \in I$. And $i \in I \subseteq L$, so $\ell + i \in L$. Then $a \in L$. Thus, $L$ is prime.
(e) follows immediately from (d). By part (d), $I$ is a maximal ideal iff $Im_\pi(I)$ is. And it's easy to show that $(0)$ is a maximal ideal of a ring exactly when the ring is a field.