= Solution
Write $\tan x=a_1x+a_3x^3+a_5x^5+O(x^7)$. Multiplying by the <cosine> <Maclaurin series> and comparing with the <sine> series gives
$$
(a_1x+a_3x^3+a_5x^5)\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7).
$$
Coefficient comparison yields $a_1=1$, $a_3-a_1/2=-1/6$ and $a_5-a_3/2+a_1/24=1/120$. Hence
$$
\boxed{\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7)}.
$$
Substitute $u=\tan x$ into $e^u=1+u+u^2/2+u^3/6+O(u^4)$. Keeping one extra order to make the term-counting explicit,
$$
e^{\tan x}=1+x+\frac{x^2}{2}+\frac{x^3}{2}+O(x^4).
$$
Counting the constant term, the first three nonzero terms are $\boxed{1+x+x^2/2}$. If the phrase is taken to mean the first three nonconstant terms, they are $x+x^2/2+x^3/2$; the displayed expansion supplies both conventions. The cubic coefficient is $1/3+1/6=1/2$, combining the cubic term in $\tan x$ with the cubic exponential term.
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