= Solution
The <Brownian first-passage time> $T_a$ is finite <almost surely>. Indeed the <Brownian reflection principle> gives $\mathbb P(T_a\leq t)=2\mathbb P(B_t\geq a)\to1$ as $t\to\infty$. Continuity of the paths means the first passage to $x+y$ occurs after the first passage to $x$. By the <Strong Markov property>, the process $B_{T_x+s}-x$ is an independent fresh <Brownian motion>. Thus $T_{x+y}-T_x$ has the same <probability distribution> as $T_y$ and is independent of $T_x$. Taking the <Laplace transforms of nonnegative random variables> gives
$$
\boxed{\varphi_{x+y}(\lambda)=\varphi_x(\lambda)\varphi_y(\lambda).}
$$
For fixed $\lambda\geq0$, these transforms lie in $(0,1]$. Put $g(a)=-\log\varphi_a(\lambda)$. It is additive on positive arguments and nondecreasing, because the hitting times increase with the level. Writing $c(\lambda)=g(1)$, additivity gives $g(r)=rc(\lambda)$ for positive rational $r$, and rational upper and lower approximations then give $g(a)=ac(\lambda)$ for every $a>0$. Hence $\varphi_a(\lambda)=e^{-ac(\lambda)}$, with $c(\lambda)\geq0$.
The <Brownian scaling> identity $T_a\overset d=a^2T_1$ gives
$$
e^{-ac(\lambda)}=\varphi_1(a^2\lambda)=e^{-c(a^2\lambda)}.
$$
For $\lambda>0$, take $a=\sqrt\lambda$ in this relation with the transform parameter one. It follows that $c(\lambda)=C\sqrt\lambda$, where $C=c(1)\geq0$. At zero, finiteness of the hitting time gives $c(0)=0$. Thus $\boxed{\varphi_a(\lambda)=e^{-aC\sqrt\lambda}}$, with the constant determined in part (b).
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