The hazard rate discussed in the question is the rate that should appear in the inhomogeneous Poisson process.
Once again we are interested in probability that the event does not happen before time $t$, which we can
determine using the equation for the probability of having $n$ events by time $t$ that was given at the top of
this page. We can thus write that the probability the default will not have occured by time $t$ as:
$$
S_T(t) = P(T>t) = \exp\left(-\int_0^t \lambda(t') \textrm{d}t' \right)
$$
Inserting the hazard rate that we are given in the question into the integral above we find that:
$$
S_T(t) = \exp\left(-\int_0^t (a + bt) \textrm{d}t' \right) = \exp\left( - at - \frac{1}{2}bt^2 \right)
$$
We now want to work out the expected value for the time of the default. As in question one above we do by first noting that the
probability that the first event will have occured by time $t$ - the cumulative probability distribution function for the
random variable $T$ - is given by:
$$
P(T \le t ) = 1 - S_T(t) = 1 - \exp\left(-\int_0^t \lambda(t) \textrm{d}t'\right)
$$
We can get the probability density function for this random variable $T$ by differentiating the above equation with respect to $t$:
$$
f_{T}(t) = \frac{ \textrm{d} P(T \le t ) }{\textrm{d}t} = \lambda(t) \exp\left(-\int_0^t \lambda(t) \textrm{d}t'\right)
$$
We can thus write the expectation of $T$, which is the quantity we are asked to calculate in the question as follows:
$$
\mathbb{E}(T) =\int_0^\infty t f_{T}(t) \textrm{d}t = \int_0^\infty t \lambda(t) \exp\left( - \int_0^t \lambda(t') \textrm{d} t' \right) \textrm{d}t
$$
It is interesting to note at this stage that we can simplify the above expression by doing a little integration by parts. Remember that the intetration
by parts formula tells us that:
$$
\int v \frac{\textrm{d}u}{\textrm{d}x} \textrm{d}x = uv - \int u \frac{\textrm{d}v}{\textrm{d}x} \textrm{d}x
$$
In this case we will set:
$$
u = t \qquad \rightarrow \qquad \frac{\textrm{d}u}{\textrm{d}t} = 1 \qquad \textrm{and} \qquad
\frac{\textrm{d}v}{\textrm{d}t} = \lambda(t) \exp\left(- \int_0^t \lambda(t') \textrm{d} t' \right)
\qquad \rightarrow \qquad v = -\exp\left(- \int_0^t \lambda(t') \textrm{d} t' \right)
$$
Inserting these results into the above formula we find that:
$$
\begin{aligned}
\int_0^\infty t \lambda(t) \exp\left( - \int_0^t \lambda(t') \textrm{d} t' \right) \textrm{d}t & = \left[ -t\exp\left(- \int_0^t \lambda(t') \textrm{d} t' \right) \right]_0^\infty +
\int_0^\infty \exp\left(- \int_0^t \lambda(t') \textrm{d} t' \right) \textrm{d}t \\
& = \int_0^\infty \exp\left(- \int_0^t \lambda(t') \textrm{d} t' \right) \textrm{d}t
\end{aligned}
$$
as the term in bracekts in the first line here is equal to zero at both limits if $\lambda(t)$ is a monotonically increasing function as it is in this question.
We will thus find the expectation in this question by solving this slightly simpler looking integral as shown below:
$$
\mathbb{E}(T) = \int_0^\infty \exp\left(-\int_0^t \lambda(t') \textrm{d}t'\right) \textrm{d}t = \int_0^\infty \exp\left(-at - \frac{1}{2} bt^2 \right) \textrm{d}t
$$
Lets simplify the expression inside the expoential here by completing the square:
$$
-at-\frac{1}{2} bt^2 = -\frac{1}{2}\left(\sqrt{b}t + \frac{a}{\sqrt{b}} \right)^2 + \frac{1}{2}\left(\frac{a}{\sqrt{b}}\right)^2
$$
When this is inserted into the integral we find:
$$
\mathbb{E}(T) = \int_0^\infty \exp\left[ -\frac{\left(\sqrt{b}t + \frac{a}{\sqrt{b}} \right)^2}{2} + \frac{a^2}{2b} \right] \textrm{d}t = \exp\left(\frac{a^2}{2b}\right) \int_0^\infty \exp\left[ -\frac{1}{2}\left(\sqrt{b}t + \frac{a}{\sqrt{b}} \right)^2 \right] \textrm{d}t
$$
We can solve the remaining integral here by making the substitution $y=\sqrt{b}t + \frac{a}{\sqrt{b}}$. Remembering that $\frac{\textrm{d}y}{\textrm{d}t} = \frac{1}{\sqrt{b}}$ here we
find that the integral we need to solve once this substitution is made is:
$$
\mathbb{E}(T) = \frac{\exp\left(\frac{a^2}{2b}\right)}{\sqrt{b}} \int_{a/\sqrt{b}}^\infty \exp\left( - \frac{y^2}{2} \right) \textrm{d}y
$$
To conclude we notice that $\frac{1}{\sqrt{2\pi}} \int_{a/\sqrt{b}}^\infty \exp\left( - \frac{y^2}{2} \right) \textrm{d}y = 1 - \Phi\left(\frac{a}{\sqrt{b}}\right)$,
where $\Phi(x)$ is the cumulative probability distribution function for a Normal distribution. We can thus write:
$$
\mathbb{E}(T) = \exp\left(\frac{a^2}{2b}\right) \sqrt{\frac{2\pi}{b}} \left[ 1 - \Phi\left(\frac{a}{\sqrt{b}}\right) \right]
$$
