Proof.
Let us first prove part (2). Since P = Z \times _ S \mathbf{P}^{n - 1}_ S we see that
\Omega ^ a_{P/S} = \bigoplus \nolimits _{a = r + s} \text{pr}_1^*\Omega ^ r_{Z/S} \otimes \text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S}
Recalling that p = \text{pr}_1 by the projection formula (Cohomology, Lemma 20.54.2) we obtain
p_*\Omega ^ a_{P/S} = \bigoplus \nolimits _{a = r + s} \Omega ^ r_{Z/S} \otimes \text{pr}_{1, *}\text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S}
By the calculations in Section 50.11 and in particular in the proof of Lemma 50.11.3 we have \text{pr}_{1, *}\text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S} = 0 except if s = 0 in which case we get \text{pr}_{1, *}\mathcal{O}_ P = \mathcal{O}_ Z. This proves (2).
By the material in Section 50.10 and in particular Lemma 50.10.4 we have \pi _*\Omega ^ a_{L/S} = \Omega ^ a_{P/S} \oplus \bigoplus _{k \geq 1} \Omega ^ a_{L/S, k}. Since the composition \pi \circ 0 in the diagram above is the identity morphism on P to prove part (3) it suffices to show that \Omega ^ a_{L/S, k} has vanishing higher cohomology for k > 0. By Lemmas 50.10.2 and 50.10.4 there are short exact sequences
0 \to \Omega ^ a_{P/S} \otimes \mathcal{O}_ P(k) \to \Omega ^ a_{L/S, k} \to \Omega ^{a - 1}_{P/S} \otimes \mathcal{O}_ P(k) \to 0
where \Omega ^{a - 1}_{P/S} = 0 if a = 0. Since P = Z \times _ S \mathbf{P}^{n - 1}_ S we have
\Omega ^ a_{P/S} = \bigoplus \nolimits _{i + j = a} \Omega ^ i_{Z/S} \boxtimes \Omega ^ j_{\mathbf{P}^{n - 1}_ S/S}
by Lemma 50.8.1. Since \Omega ^ i_{Z/S} is free of finite rank we see that it suffices to show that the higher cohomology of \mathcal{O}_ Z \boxtimes \Omega ^ j_{\mathbf{P}^{n - 1}_ S/S}(k) is zero for k > 0. This follows from Lemma 50.11.2 applied to P = Z \times _ S \mathbf{P}^{n - 1}_ S = \mathbf{P}^{n - 1}_ Z and the proof of (3) is complete.
We still have to prove (1). If n = 1, then we are blowing up an effective Cartier divisor and b is an isomorphism and we have (1). If n > 1, then the composition
\Gamma (X, \Omega ^ a_{X/S}) \to \Gamma (L, \Omega ^ a_{L/S}) \to \Gamma (L \setminus E, \Omega ^ a_{L/S}) = \Gamma (X \setminus Z, \Omega ^ a_{X/S})
is an isomorphism as \Omega ^ a_{X/S} is finite free (small detail omitted). Thus the only way (1) can fail is if there are nonzero elements of \Gamma (L, \Omega ^ a_{L/S}) which vanish outside of E = 0(P). Since L is a line bundle over P with zero section 0 : P \to L, it suffices to show that on a line bundle there are no nonzero sections of a sheaf of differentials which vanish identically outside the zero section. The reader sees this is true either (preferably) by a local calculation or by using that \Omega _{L/S, k} \subset \Omega _{L^\star /S, k} (see references above).
\square
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