Moral Hazard with risk neutral agent
A thing that bothers me here is the following: the Incentive Compatibility constraint is 

$$IC: w'p(a') + w(1-p(a')) - 1 \geq w'p(a) + w(1-p(a))$$

$$\implies w'-w \geq \frac {1}{p(a')-p(a)}  \tag{1}$$

...since by assumption $p(a')-p(a) &gt;0$. We are told that we should find that at the optimum, 
$$x'-w' = x-w \implies x'-x = w'-w \tag{2}$$

Combining $(1)$ and $(2)$, if indeed this is the optimum under the given constraints, we must also have

$$x'-x \geq \frac {1}{p(a')-p(a)} \tag{3}$$

But this is an additional, necessary constraint on a priori magnitudes, that must hold if the postulated optimal solution is to be admissible. Even if indeed such a constraint is assumed, in any case, it visibly reduces the generality of the problem (which purports to show something general, i.e. how the risk-neutrality of the agent affects the solution). 

Nevertheless, let's work this a bit more formally. I will assume that $w, w'$ can be zero, but not negative. This is a maximization problem in normal form with inequality constraints, non-negative decision variables and non-negative multipliers. The full Lagrangean of the problem therefore is (I will compact notation in an obvious way),

$$\Lambda =  u(x'-w')p' + u(x-w)(1-p') + \lambda\cdot [w'p' + w(1-p') - 1 ]\\ + \mu \cdot [ w'p' + w(1-p') - 1 - w'p - w(1-p)] + \xi w + \xi' w'$$

The essential first order conditions are

$$\frac {\partial \Lambda}{\partial w} \leq 0, \;\;\frac {\partial \Lambda}{\partial w} \cdot w = 0$$

and analogously for $w'$. These result in

$$\frac {\partial \Lambda}{\partial w} = -u'(x-w)(1-p') +\lambda (1-p') - \mu (p'-p) + \xi \leq 0$$

$$\implies u'(x-w)(1-p') \geq \lambda (1-p') - \mu (p'-p) + \xi $$

$$\implies u'(x-w) \geq \lambda - \mu \frac {p'-p}{1-p'} + \frac {\xi}{1-p'} \tag{4}$$

$$\frac {\partial \Lambda}{\partial w'} = -u'(x'-w')p' +\lambda p' + \mu (p'-p) + \xi' \leq 0$$

$$\implies u'(x'-w') \geq \lambda + \mu\frac {p'-p}{1-p'} + \frac {\xi'}{p'} \tag{5}$$

First note that not both wages can be zero, because the constraints would be violated. Given this, consider the possibility that the $IR$ is binding (so $\lambda &gt;0$). If it is binding, then with not both wages zero, the $IC$ constraint will necessarily be violated. So we conclude that

$$\lambda_* = 0$$

and the first-order conditions now become

$$u'(x-w) \geq  - \mu \frac {p'-p}{1-p'} + \frac {\xi}{1-p'} \tag {4a}$$

$$ u'(x'-w') \geq  \mu \frac{p'-p}{1-p'} + \frac {\xi'}{p'} \tag{5a}$$

Now note that if $\xi =0$ (i.e. $w &gt;0$) then $(4a)$ should hold as an equality and with the last term on the right equal to zero. But this would require negative marginal utility which is inadmissible. We also know that not both wages can be zero. So we conclude that we must have

$$\xi_* &gt; 0 , w_*=0,\;\;\; \xi'_* = 0,\; w'_* &gt;0$$

and the conditions now become

$$u'(x) \geq  - \mu \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'} \tag {4b}$$

$$ u'(x'-w') =  \mu \frac{p'-p}{1-p'}  \tag{5b}$$

Eq. $(5b)$ implies that $\mu_*&gt;0$ , under a usual utility function specification, which does not give zero marginal utility except at infinity. This in turn means that the $IC$ constraint should hold as an equality. Given that $w_*=0$ this gives

$$IC:  w'p'  - 1 - w'p =0 \implies = w'_* = \frac 1{p'-p} \tag{6}$$ 

This should ring a bell, because the right-hand-side of $(6)$ is the same as the right-hand-sides of $(1)$ and $(3)$.

Namely, if we are assuming a priori that $x'-x = \frac 1{p'-p}$, then the solution we have arrived at validates the claim $x'-w'_* = x-w_*$

Under this additional assumption, we also obtain

$$u'(x) \geq  - \mu_* \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'} \tag {4c}$$

$$ u'(x) =  \mu_* \frac{p'-p}{1-p'}  \tag{5c}$$

Combining, we obtain

$$ \mu \frac{p'-p}{1-p'} \geq - \mu \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'}$$

$$\implies \mu_* \geq \frac {\xi_*}{2(p'-p)} \tag{7}$$

This is admissible. So under $x'-x = \frac 1{p'-p}$, we obtain the solution 

$$\left \{w'_* = x'-x = 1/(p'-p), w_* =0, \lambda_*=0, \mu_* \geq \frac {\xi_*}{2(p'-p)}, \xi_* &gt;0, \xi'_* =0\right\}$$