Computation of $\operatorname{Tor}_1$ and $\operatorname{Ext}^1$ Can you please give some examples of computation of the derived functors $\operatorname{Tor}_1$ and $\operatorname{Ext}^1$ for some simple cases, say $R=\mathbb{Z}$ or $R=\mathbb{Z}[G]$ for some finite group $G$?
 A: Here is a slightly more complicated example than just free abelian groups. We will compute $\textrm{Ext}_{\Bbb{Z}/4}^1(\Bbb{Z}/2,\Bbb{Z}/2)$. 
Consider $\Bbb{Z}/2$ as a $\Bbb{Z}/4$ - module; the universal property of quotients gives us a map (call it $g_0$) from $\Bbb{Z}/4 \to \Bbb{Z}/2$. Then we get by continuing a similar process a free resolution of $\Bbb{Z}/2$ by free $\Bbb{Z}/4$ modules
$$\ldots \stackrel{g_3}{\longrightarrow} (\Bbb{Z}/4)^8 \stackrel{g_2}{ \longrightarrow} (\Bbb{Z}/4)^2 \stackrel{g_1}{ \longrightarrow } \Bbb{Z}/4 \stackrel{g_0}{\longrightarrow} \Bbb{Z}/2 \longrightarrow 0$$
where the map $g_1$ sends $(1,0)$ to $0$ and $(0,1)$ to $2$, $g_2$ sends each canonical generator to an element of the kernel of $g_1$ and so on. Now we take $\textrm{Hom}(-,\Bbb{Z}/2)$ (where our homs are now $\Bbb{Z}/4$ - homs) of this exact sequence to get the chain complex
$$\ldots  \stackrel{g_3^\ast}{\longleftarrow} \textrm{Hom}((\Bbb{Z}/4)^8 ,\Bbb{Z}/2) \stackrel{g_2^\ast}{ \longleftarrow} \textrm{Hom}( (\Bbb{Z}/4)^2 ,\Bbb{Z}/2) \stackrel{g_1^\ast} { \longleftarrow } \textrm{Hom}(\Bbb{Z}/4,\Bbb{Z}/2) \stackrel{g_0^\ast}{\longleftarrow} \textrm{Hom}(\Bbb{Z}/2,\Bbb{Z}/2)  \\ \hspace{5.5in}\longleftarrow 0$$
Now firstly we have $\textrm{Ext}_{\Bbb{Z}/4}^0(\Bbb{Z}/2,\Bbb{Z}/2)$ being isomorphic to $\textrm{Hom}(\Bbb{Z}/2,\Bbb{Z}/2) \cong \Bbb{Z}/2.$
To compute $\textrm{Ext}^1$, first notice that $g_1^\ast$ is the zero map. For if $f : \Bbb{Z}/4 \to \Bbb{Z}/2$, precomposing it with $g_1$ gives that
$$f \circ g_1= 0$$
because the image of $g_1$ is the two point set $\{0,2\}$. Any $f :\Bbb{Z}/4 \to \Bbb{Z}/2$ evaluated on this set is zero. It remains to determine the kernel of $g_2^\ast$.
Now  notice that each of the homs is isomorphic to a direct sum of $\Bbb{Z}/2$'s. In the case of computing the kernel of $g_2^\ast$, we get that $g_2^\ast$ is a map from $\textrm{Hom}((\Bbb{Z}/4)^2,\Bbb{Z}/2) \cong \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^2$ to 
$$\textrm{Hom}((\Bbb{Z}/4)^8,\Bbb{Z}/2) \cong \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^8.$$
We now just need to compute the kernel of the associated map $$h : \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^2 \to \textrm{Hom}((\Bbb{Z}/4),\Bbb{Z}/2)^8.$$
The kernel of this map consists of the 0 tuple and the tuple $(\varphi,\varphi)$ where $\varphi : \Bbb{Z}/4 \to \Bbb{Z}/2$ that sends $1$ to $1$. 
It follows that
$$\textrm{Ext}_{\Bbb{Z}/4}^1(\Bbb{Z}/2,\Bbb{Z}/2) \cong \Bbb{Z}/2.$$
A: Here's a simple example of $\mathrm{Tor}_n^R$ for $R = \mathbb Z$:
Example 1: $\mathrm{Tor}_n^{\mathbb Z} (\mathbb Z / 2 \mathbb Z, \mathbb Z / 2 \mathbb Z)$
(i) Choose a projective (in this case free) resolution of $\mathbb Z / 2 \mathbb Z$:
$$ \dots \to 0 \to \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \xrightarrow{\pi} \mathbb Z / 2 \mathbb Z \to 0$$
(ii) Remove $M = \mathbb Z / 2 \mathbb Z$ and apply $- \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z$ to get
$$ 0 \to \mathbb Z \otimes_{\mathbb Z} \mathbb Z/ 2 \mathbb Z \xrightarrow{id \otimes (\cdot 2)} \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \to 0$$
Then simplify using $R \otimes_R M \cong M$ to get 
$$ 0 \to \mathbb Z / 2 \mathbb Z \xrightarrow{f} \mathbb Z / 2 \mathbb Z \to 0$$
Find $f$ for example, by using the isomorphism $R \otimes_R M \cong M , r \otimes m \mapsto rm $:
$$ \mathbb Z / 2 \mathbb Z \to \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \xrightarrow{id \otimes (\cdot 2)} \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z \to \mathbb Z / 2  \mathbb Z$$
where we have the maps $m \mapsto 1 \otimes m, 1 \otimes m \mapsto 1 \otimes 2m , 1 \otimes 2m \mapsto 2m$, in this order.
Hence we see that $f \equiv 0$.
(iii) Hence we get
$$ \mathrm{Tor}_0^{\mathbb Z} (\mathbb Z/2 \mathbb Z ,  \mathbb Z/ 2 \mathbb Z ) = \mathbb Z / 2  \mathbb Z\otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z$$
$$ \mathrm{Tor}_n^{\mathbb Z} (\mathbb Z / 2 \mathbb Z, \mathbb Z / 2  \mathbb Z) = 0 \text{ for } n \geq 2$$
$$ \mathrm{Tor}_1^{\mathbb Z}(\mathbb Z / 2  \mathbb Z,  \mathbb Z/ 2 \mathbb Z ) = \mathrm{ker}0 / \mathrm{im}f = \mathbb Z / 2 \mathbb Z$$
A: Simple example: $\text{Ext}_{Z}^{i}(Z/2,Z)$
All references in this post refer to Introduction to homological algebra, Rotman
Note that by Theorem 6.67,  $\text{Ext}_{Z}^{i}(Z/2,Z) \cong \text{ext}_{Z}^{i}(Z/2,Z)$, hence we have two ways to calculate this.
First way:
Take $\text{Ext}_{Z}^{i}(Z/2,Z)$, calculate the injective resolution of Z, we get $0 \rightarrow Z \rightarrow Q \rightarrow Q/Z \rightarrow 0 \rightarrow ...$. Note that Q and Q/Z are injective Z-module since Z is PID (Prop 3.34).
Next, apply functor $Hom_Z(Z/2,-)$ to the deleted injective resolution, we have $0 \rightarrow Hom(Z/2,Q) \rightarrow Hom(Z/2,Q/Z) \rightarrow 0 \rightarrow ...$
Hence $\text{Ext}_{Z}^{1}(Z/2,Z) \cong Hom_Z(Z/2,Z)$ by Theorem 6.45
$\text{Ext}_{Z}^{n}(Z/2,Z) \cong 0$ if $n \geq 2$
You can check that $Hom(Z/2,Q) \cong 0$ and $ Hom(Z/2,Q/Z) \cong Z/2Z$
Hence $\text{Ext}_{Z}^{n}(Z/2,Z) \cong Z/2Z$
Second way:
Take $\text{ext}_{Z}^{i}(Z/2,Z)$,
Take projective resolution of Z/2, $... 0\rightarrow Z \rightarrow^{f} Z \rightarrow Z/2 \rightarrow 0, f: 1 \rightarrow 2$
Then apply functor $Hom_Z(-,Z)$ to the deleted resolution, we have
$0 \rightarrow Hom(Z,Z) \rightarrow Hom(Z,Z) \rightarrow 0 \rightarrow ...$
Hence we have the same results.
