= Solution
For $0<b<\lambda=2$, integrate the identity $e^{bz}-1=\int_0^z be^{bt}\,dt$ for $z\geq0$. The <Tonelli theorem> and the tail bound in (c) give
$$
\mathbb E e^{bZ}=1+b\int_0^\infty e^{bt}\mathbb P(Z>t)\,dt
\leq1+b\int_0^\infty e^{-(2-b)t}\,dt=\frac2{2-b}.
$$
For $b=0$ the <expectation> is $1$, and for $b<0$ the fact $Z\geq0$ gives $0<e^{bZ}\leq1$. Therefore
$$
\boxed{\mathbb E e^{bZ}<\infty\quad\text{for every }b<2.}
$$
This is an <exponential moment bound for a Gaussian random-walk maximum>. The upper-tail bound proves the requested sufficient range; it alone does not establish a converse at the endpoint.
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