I am trying to learn computations of the ASS by myself from "user's guide of spectral sequences" book and here is the thing I want to compute:
Compute $\operatorname{Ext}_{\mathcal{A}_2}^{s,t}(\mathcal{A}_2/ I(\mathcal A_2 . Sq^1), \mathbb F_2).$
Could someone tell me what exactly are the general steps of this computation?
In my opinion, the exercise is asking the wrong question: either that or the notation is very strange. First, this exercise seems to be related to the proof of Corollary 9.48, and so I think that the exercise has a typo: it intends to ask about $\mathcal{A}_2/(I(\mathcal{A}_2) \cdot Sq^1)$, where $I(\mathcal{A}_2)$ should, if the notation is sane, denote the augmentation ideal of $\mathcal{A}_2$: the ideal of all elements in positive degree, which is also the same as the kernel of the augmentation map $\mathcal{A}_2 \to \mathbb{F}_2$. Then $I(\mathcal{A}_2) \cdot Sq^1$ consists of all elements of the form $a Sq^1$ where $a \in I(\mathcal{A}_2)$. I think that in the proof of Corollary 9.48 (and hence in this exercise), it should instead say $\mathcal{A}_2 \cdot Sq^1$, the left ideal generated by $Sq^1$. This has the properties used in the proof, whereas $I(\mathcal{A}_2) \cdot Sq^1$ does not; in particular, I believe that $I(\mathcal{A}_2) / I(\mathcal{A}_2) \cdot Sq^1$ is neither $1$-connected nor free over $A(0)$ (top of p. 434). (Another point is, you want $Sq^1$ to lie in the object that McCleary is denoting $I(\mathcal{A}_2) \cdot Sq^1$, and it doesn't: there is no positive degree element $a \in \mathcal{A}_2$, which is to say no element $a \in I(\mathcal{A}_2)$, such that $Sq^1 = a Sq^1$.)
In short, I think that the exercise should ask about the same thing used in the proof. Exercise: compute $\mathrm{Ext}^{s,t}_{\mathcal{A}_2}(\mathcal{A}_2/(\mathcal{A}_2 \cdot Sq^1), \mathbb{F}_2)$.
Sketch: show that $\mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1) = \mathcal{A}_2 \otimes_{A(0)} \mathbb{F}_2$ and then use Fact 3 (change-of-rings) on p. 438. Alternatively, once you show that $\mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1) = \mathcal{A}_2 \otimes_{A(0)} \mathbb{F}_2$, then you can construct an explicit projective resolution of $\mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1)$ as an $\mathcal{A}_2$-module. This is probably the approach McCleary is alluding to at the bottom of p. 433. You can do this by starting with the short exact sequence of $A(0)$-modules $$ 0 \to \Sigma \mathbb{F}_2 \to A(0) \to \mathbb{F}_2 \to 0 $$ (where $\Sigma \mathbb{F}_2$ just means $\mathbb{F}_2$ in degree 1). Apply the exact functor $\mathcal{A}_2 \otimes_{A(0)} -$ to get $$ 0 \to \Sigma \mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1) \to \mathcal{A}_2 \to \mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1) \to 0. $$ Now since the kernel and cokernel are isomorphic up to a shift in dimensions, you can use this to get a nice free resolution of $\mathcal{A}_2 / (\mathcal{A}_2 \cdot Sq^1)$.