Double implication The present question is about a double implication. 
I have the following implications:
$(a)\implies (b)\implies ((c)$ or $(a))$
Can I deduce that $(b)\implies(a)$ only?
 A: 
$(a)\implies (b)\implies ((c)$ or $(a))$
Can I deduce that $(b)\implies(a)$ only?

No.
What you have is the following:
$$(a) \implies (b)\tag{1}$$
$$(b)\implies (c\lor a) \equiv \lnot b \lor c \lor a\tag{2}$$
In (2) I use the equivalence $p\implies q \equiv \lnot p \lor q$
From (2), It may very well have that each $\lnot b$ is true, and $a$ is true, and $c$ is true. It may very well be that we have only $\lnot b$ is true, or only $a$ is true, or only $c$ is true. Or it may very well be that we have only $\lnot b\lor c\equiv b\implies c$ is true. Or, we may have that $\lnot b \lor a \equiv b\implies a$ is true. Or both these last cases. 
In short, $b\implies a$ is satisfiable (possible), but it is not logically necessary/valid. 
\begin{array}{a|b|c|aVc|rem|}
\hline
a&b&c&a\vee c& \\
\hline
F&F&F&F&\\
F&F&T&T\\
F&T&F&F\\
F&T&T&T& \text{Here, b $\rightarrow (a \lor c)$ is true, but $b \rightarrow a$ is false.}\\
T&F&F&T\\
T&F&T&T\\
T&T&F&T\\
T&T&T&T\\
\end{array}
There's no logical justification for concluding it must be the case that $b\implies a$. 
**However, if you know that $a \equiv \lnot c$, then that counterexample is ruled out, because in the above indicated counterexample, we have $a = c = T$.

Now, if you also have the premise $$\lnot c\tag{3},$$ then $(3)$ together with $(2)$ imply that $\lnot b \lor a \equiv b\implies a$, by the disjunctive syllogism $\lnot b \lor c \lor a \equiv (\lnot b \lor a) \lor c$:
$\;\;\;(\lnot b \lor a) \lor c$
$\;\;\;\lnot c$
$\therefore (\lnot b \lor a)\equiv b\implies a$.
A: See a truth table. It is often helpful.
$$
\begin{array}{a|b|c|aVc|rem|}
\hline
a&b&c&a\vee c& \\
\hline
F&F&F&F&\\
F&F&T&T\\
F&T&F&F\\
F&T&T&T& \text{Here, b $\implies$a $\vee$ c is true but b $\implies$a is not.}\\
T&F&F&T\\
T&F&T&T\\
T&T&F&T\\
T&T&T&T\\
\end{array}
$$
Note that False can imply Truth but truth cannot imply False.
So your statement is SATISFIABLE but not VALID.
$\textbf{Edit}$ Edited in response to some silly mistake.
A: Although truth tables provide a systematic approach to problems like this, it's often more efficient to just think about "what can go wrong?" --- in this case, how (if at all) could the  assumptions $(a\implies b)$ and $(b\implies(c\lor a))$ be true while the proposed conclusion $(b\implies a)$ is false?  Falsehood of an implication is quite restrictive; it requires both that the antecedent is true and that the consequent is false, so in this case, to falsify the proposed conclusion, you'd need $b$ true and $a$ false.  That automatically makes the first hypothesis, $(a\implies b)$, true.  As for the second hypothesis, its antecedent $b$ is true, so the only way the implication could be true is for the consequent to be true.  In this case, you'd need $(c\lor a)$ to be true.  Since $a$ is false, you'd need $c$ true.  In this way you find a truth assignment, "$a$ false and both $b$ and $c$ true," that makes your hypotheses both true and the proposed conclusion false, so the inference is not correct.
