I managed to solve a) by differentiation. But, I am stuck at b). I can't see why change from $B$ to $e^B$ puts the series of a) in the exponential function. Also, I think I have to use a) and b) to show c). However I can't find a way to do so. Could anyone please explain to me?
Ok here we go, using the identities given we clearly have $$ \begin{align} e^{sA}e^{sB} &= e^{sA} e^{sB} e^{-sA} e^{sA} \\ &= \exp\left(sB + s^{2}[A,B] + \frac{s^3}{2}[A,[A,B]] + \mathcal{O}(s^4) \right)e^{sA} \tag{1} \end{align} $$ and we want to show that after expansion this agrees, up to terms of $\mathcal{O}(s^4)$, with $$ \exp\left(s(A+B) + \frac{s^2}{2}[A,B] + \frac{s^3}{12}\left([[A,[A,B]] + [B,[B,A]]\right) + \mathcal{O}(s^4)\right) \tag{2} $$ the goal is just to expand both of these equations, collect terms and then show they agree.
Worked example for the $s^2$ term
I will demonstrate for terms involving $s^2$ and let you try the others. So first define the argument of the exponential in $(2)$ by $$ Z = s(A+B) + \frac{s^2}{2}[A,B] + \frac{s^3}{12}[[A,[A,B]] + [B,[B,A]] $$ expanding the power series for the exponential and keeping only those terms involving $s^2$ we have $$ I + Z + \frac{Z^2}{2} + \cdots \mapsto \frac{s^2}{2}[A,B] +\frac{s^2}{2}(A^2 + AB + BA + B^2) \tag{3}. $$ So after defining $$ X = sB + s^2[A,B] + \frac{s^3}{2}[A,[A,B]] + \mathcal{O}(s^4) $$ the hope is that after similarly expanding $e^X e^{sA}$ and collecting terms the two polynomials agree. Now noting that we only need to consider terms up to $X^2$ consider \begin{align} \left( I + X + \frac{X^2}{2} \right)\left(I + sA + \frac{s^2A^2}{2} \right) \tag{4} \end{align} quick checks tell us that $$ X^2 = s^2B^2 +s^3B[A,B] +s^3[A,B]B+\mathcal{O}(s^4) $$ so after carrying out the multiplication (4) and collecting only those terms involving $s^2$ we get $$ \begin{align*} \frac{s^2 A^2}{2} + s^2 B A + s^2[A,B] + \frac{s^2 B^2}{2} &= \frac{s^2}{2}[A,B] + \frac{s^2}{2} \left([A,B] + 2 BA + A^2 + B^2 \right) \\ &= \frac{s^2}{2}[A,B] + \frac{s^2}{2}\left(A^2 + AB + BA + B^2 \right) \end{align*} $$ which as hoped tells us that those terms with coefficients $s^2$ of both identities agree.
I shall leave it to you to repeat the process for $s^0, s^1$ and $s^3$ applying the same ideas, hope that has helped.