One more example refering to the second question: Let $$\Gamma = \lbrace left\lbrace \left. \begin{pmatrix} 1 & \ast & \ast \newline & 1 & \ast \newline & & 1 \end{pmatrix} \mid right\vert\ \ast \in \mathbb{Z} \rbrace$$ right\rbrace$$ be the group of integral upper triangular matrices with unit diagonal. $\Gamma$ fits into the non-split central extension $$0 \to \mathbb{Z} \to \Gamma \to \mathbb{Z}^2 \to 0$$ that corresponds to a generator $\epsilon \in H^2(\mathbb{Z}^2;\mathbb{Z}) = \mathbb{Z}$.
Claim 1: $cd(\Gamma) = 3$
Since $\mathbb{Z}$ resp. $\mathbb{Z}^2$ has cd $1$ resp. $2$, the LHS spectral sequence $E_2^{ij} = H^i(\mathbb{Z}^2;H^j(\mathbb{Z};M))$ shows $cd(\Gamma) \le 3$. Moreover, by positional reasons $E_\infty^{2,1}=E_2^{2,1}$ and in particular $E_\infty^{2,1} = \mathbb{Z}$ for $M = \mathbb{Z}$. Hence $cd(\Gamma) = 3$.
Claim 2: The inflation map $\text{inf}: H^2(\mathbb{Z}^2;M) \to H^2(\Gamma;M)$ is zero.
Since the image of inflation is just $E_\infty^{2,0} = \text{coker}(d_2^{0,1})$, it is sufficient to show that $d_2^{0,1}$ is surjective. Since the action of $\Gamma$ on $M$ is induced by the action of $\mathbb{Z}^2$, it follows that $\mathbb{Z}$ acts trivially on $M$. Futhermore, $\mathbb{Z}^2$ acts trivially on $H^\ast(\mathbb{Z};M)$ because $\mathbb{Z}$ is central. Therefore $E_2^{0,1} = Hom(\mathbb{Z},M)$.
Let $\alpha \in Hom(\mathbb{Z},M)$. Then $$d_2^{0,1}: Hom(\mathbb{Z},M) \to H^2(\mathbb{Z}^2;M)$$ is given by $d_2(\alpha)= - \alpha^\ast(\epsilon)$ (well-known formula) where $$\alpha^\ast: H^2(\mathbb{Z}^2;\mathbb{Z}) \to H^2(\mathbb{Z}^2;M)$$ is induced by $\alpha$ on the coefficients. By using a projective resolution or by Poincare duality one easily sees that $$\alpha^\ast : \mathbb{Z} \to M/\lbrace gm-m \mid g \in \mathbb{Z}^2 \rbrace =: \bar{M}$$ is just $\alpha$ composed with the natural projection. Identifying $Hom(\mathbb{Z},M) = M$ now shows that $d_2^{0,1}: M \to \bar{M}$ is the natural projection and hence surjective.
