Question on uncertainity
We are given that $u$ is increasing and concave, and $u(0) = 0$. This implies that $\dfrac{u(t)}{t}$ is decreasing in $t$, and also, $\dfrac{u(t)}{t} &gt;  u'(t)$ for all $t$.

Nicole's maximization problem :
\begin{eqnarray*} \max_{x} \  q(x)(w-t(x))\end{eqnarray*}

FOC :
$q'(x)(w-t(x)) = q(x)t'(x)$

Suppose $x_N$ solves Nicole's problem. Therefore, it satisfy

$q'(x_N)(w-t(x_N)) = q(x_N)t'(x_N)$
or equivalently,

$\dfrac{q'(x_N)}{q(x_N)t'(x_N)} = \dfrac{1}{w-t(x_N)} \tag{1}$

Andrew's maximization problem :
\begin{eqnarray*} \max_{x} \  q(x)u(w-t(x))\end{eqnarray*}

FOC :
$q'(x)u(w-t(x)) = q(x)t'(x)u'(w-t(x))$

or equivalently,

$\dfrac{q'(x)u(w-t(x))}{q(x)t'(x)} = u'(w-t(x)) \tag{2}$

Let us compare the two sides of the above FOC (2) at $x = x_N$,

LHS  = $\dfrac{q'(x_N)u(w-t(x_N))}{q(x_N)t'(x_N)} = \dfrac{u(w-t(x_N))}{w-t(x_N)} &gt; u'(w-t(x_N)) =$ RHS

Since at $x = x_N$ LHS of (2) is greater than RHS of (2), we need to reduce the value of $x$ to make them equal. So, there is $x_A &lt; x_N$ such that 

$\dfrac{q'(x_A)u(w-t(x_A))}{q(x_A)t'(x_A)} = u'(w-t(x_A)) \tag{3}$

Therefore, Nicole buys larger quantity of technology. 

Observing the FOCs, we can also conclude that increase in $w$ lead to increase in optimal quantity chosen by both.

Now try to solve the remaining two parts.