The Bianchi identity is $\triangledown_{[aR_{bc}]_d^e}=0$.
Contraction of the Bianchi identity leads to an important equation satisfied by $R_{ab}$. We find
$$\triangledown_a R_{bcd}^a + \triangledown_bR_c^b-\triangledown_cR_{bd}=0$$
Raising the index $d$ with the metric and contracting over b and d, we obtain
$$\triangledown_aR_c^a+\triangledown_bR_c^b - \triangledown_c R = 0$$ or
$$\triangledown^aG_{ab}=0$$
where
$$G_{ab}=R_{ab}-\frac{1}{2}Rg_{ab}$$
1) How do you get equation one?
2) What do you they mean by raising the index and contracting over b and d and why can we do this?
3) How do we get the last equation from the second from last one?
This is from Wald's General Relativity, page 40.
I think you're a little confused, possibly due to a misprint in Wald. We should have $$\nabla^a R_{abcd}+\nabla_b R_{cd}-\nabla_c R_{bd}=0.$$(I recommend you carefully read whatever proof Wald provides or, if he doesn't give one, try it as an exercise given his hint in terms of the Bianchi identity.) Multiplying by $g^{bd}$ gives $$\nabla^a R_{ac}+\nabla^d R_{cd}-\nabla_c R=0.$$As the Ricci scalar is symmetric, this may be rewritten as $$2\nabla^a R_{ac}-\nabla_c R=0.$$The definition $G_{ab}:=R_{ab}-\frac{R}{2}g_{ab}$ gives $$2\nabla^a G_{ac}=\nabla^a R_{ac}-\frac{1}{2}g_{ac}\nabla^a R=\nabla^a R_{ac}-\frac{1}{2}\nabla_c R=0.$$