Since $B_t \to \infty$ a.s. as $t \to \infty$, we conclude from the intermediate value theorem that $T<\infty$ a.s. Moreover, by the continuity of the sample paths,
$$\mathbb{E}B_T = \mathbb{E}(e^{-\lambda T}) \neq 0$$
and therefore Wald's identities imply that $T$ is not integrable (if $T$ would be integrable, then $\mathbb{E}B_T = 0$).
Concerning the Ornstein-Uhlenbeck process: Let
$$X_t = \sigma \cdot e^{b \, t} \cdot \int_0^t e^{-b \, s} \, dB_s$$
where $\sigma>0$. Then
$$X_t \geq e^{-\lambda \, t} \Leftrightarrow M_t := \int_0^t e^{-b \, s} \, dB_s \geq \frac{1}{\sigma} e^{-(\lambda+b) \, t}$$
$M$ is a martingale, $M_0 =0$. A similar calculation as in the proof of Wald's identity shows that $\mathbb{E}M_T = 0$ for any integrable stopping time $T \in L^1$. With the same reasoning as above, we conclude $T \notin L^1$.
For $c=0$ this result is knows as reflection principle (see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 6) and follows from the Markov property and symmetry of Brownian motion. However, for $c>0$ the proof is more involved since we have to get rid of the drift term.
Since by definition
$$[H_a \leq t] = \left[ \sup_{s \leq t} X_s \geq a \right] \tag{1}$$
determining the distribution of $H_a$ is equivalent to finding the distribution of $\sup_{s \leq t} X_s$. In order to find the distribution of the latter, we need two ingredients: Girsanov's theorem and the joint distribution $(X_t,\sup_{s \leq t} X_s)$ for a Brownian motion $(X_t)_{t \ge 0}$.
Girsanov theorem: Let $c \in \mathbb{R}$ and $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Then $$X_t := B_t+ct, \qquad t \leq T,$$ is a Brownian motion with respect to the probability measure $$d\mathbb{Q} := d\mathbb{Q}_T := \exp \left( -c B_T - \frac{c^2}{2} T \right) d\mathbb{P}.$$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 18.
Joint distribution of $(X_t, \sup_{s \leq t} X_s)$: Let $(X_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{Q})$. Then the joint distribution $(X_t,\sup_{s \leq t} X_s)$ equals $$\mathbb{Q} \left[ X_t \in dx, \sup_{s \leq t} X_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. \tag{2}$$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web).
So let's finally put it all together: It follows from $(1)$ and the definition of the probability measure $\mathbb{Q}_T$ that
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \mathbb{P} \left[ \sup_{s \leq T} X_s \geq a \right] = \int 1_{[a,\infty)} \left( \sup_{s \leq T} X_s \right) \, d\mathbb{P} \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right) \exp \left( c B_T + \frac{c^2}{2} T \right) \, d\mathbb{Q}_T \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right)\exp\left(c X_T- \frac{c^2}{2} T \right) \, d\mathbb{Q}_T. \end{align*}$$
By Girsanov's theorem, $(X_t)_{t \leq T}$ is a Brownian motion with respect to $\mathbb{Q}_T$ and therefore $(2)$ gives
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \exp\left(- \frac{c^2}{2} T \right) \int_{y \geq a} \int_{x \leq y} e^{cx} \frac{2 (2y-x)}{\sqrt{2\pi T^3}} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dx \, dy. \end{align*}$$
It remains to calculate the integral expression. First of all, by Fubini's theorem,
$$\begin{align*} \mathbb{P}[H_a \leq T] &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \left(\int_{x \geq a} e^{cx} I_1(x) \, dx + \int_{x \leq a} e^{-cx} I_2(x) \, dx \right) \\ &:=J_1+J_2 \tag{3} \end{align*}$$
where
$$\begin{align*} I_1(x):= \int_{y \geq x} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &= \left[ - \exp \left(- \frac{(2y-x)^2}{2T} \right) \right]_{y=x}^{\infty} \\ &= \exp \left(- \frac{x^2}{2T} \right) \\ I_2(x) := \int_{y \geq a} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &=\exp \left(- \frac{(2a-x)^2}{2T} \right). \end{align*}$$
Hence,
$$\begin{align*} J_1 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \geq a} e^{cx} I_1(x) \, dx \\ &= \frac{1}{\sqrt{2\pi T}} \int_{x \geq a} \exp \left(- \frac{(x-cT)^2}{2T} \right) \, dx \\ &= \frac{1}{\sqrt{\pi}} \int_{z \geq \frac{a-cT}{\sqrt{2T}}} \exp(-z^2) \, dz \tag{4} \end{align*}$$
and
$$\begin{align*} J_2 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \leq a} e^{cx} I_2(x) \, dx \\ &\stackrel{u:=2a-x}{=} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{u \geq a} e^{c(2a-u)} \exp\left(-\frac{u^2}{2T} \right) \, du \\ &=\ldots = \frac{e^{2ac}}{\sqrt{\pi}} \int_{z \geq \frac{a+CT}{\sqrt{2T}}} \exp(-z^2) \, dz. \tag{5} \end{align*}$$
Now if we differentiate $(3)$ with respect to $T$, using $(4)$ and $(5)$, we get
$$\begin{align*} \frac{d}{dT} \mathbb{P}(H_a \leq T) &= \frac{-1}{\sqrt{\pi}} \exp \left( - \frac{(a-cT)^2}{2T} \right) \left( \frac{-c}{\sqrt{2T}} - \frac{a-cT}{2 \sqrt{2} T^{3/2}} \right) \\ &\quad + \frac{-1}{\sqrt{\pi}} \underbrace{e^{2ac} \exp \left( - \frac{(a+cT)^2}{2T} \right)}_{\exp(-(a-cT)^2/2T)} \left( \frac{c}{\sqrt{2T}} - \frac{a+cT}{2 \sqrt{2} T^{3/2}} \right) \\ &= \frac{a}{\sqrt{2\pi T^3}} \exp \left(- \frac{(a-cT)^2}{2T} \right). \end{align*}$$
Best Answer
Your CDF is slightly wrong $P(T_a \le t) = 2 P(B_t \ge a) = 2 \Phi(-a/\sqrt{t})$.
Applying the Fundamental Theorem of Calculus yields $$ \frac{d}{dt} \left[2\int_{-\infty}^{-a/\sqrt{t}} \phi(u) \, du\right] = 2 \phi(-a/\sqrt{t}) \frac{d}{dt} (-a/\sqrt{t}) = at^{-3/2} \phi(-a/\sqrt{t}). $$