= Solution
First expand the result of the first gate in the <computational basis>:
$$
U(\alpha)|+\rangle
=\frac{|+\rangle+e^{-i\alpha}|-\rangle}{\sqrt2}
=\frac{1+e^{-i\alpha}}2|0\rangle+\frac{1-e^{-i\alpha}}2|1\rangle.
$$
Using $1+e^{-i\alpha}=2e^{-i\alpha/2}\cos(\alpha/2)$ and $1-e^{-i\alpha}=2ie^{-i\alpha/2}\sin(\alpha/2)$, this becomes
$$
e^{-i\alpha/2}\left(\cos(\alpha/2)|0\rangle+i\sin(\alpha/2)|1\rangle\right).
$$
Apply $U(\beta)|0\rangle=|+\rangle$ and $U(\beta)|1\rangle=e^{-i\beta}|-\rangle$ to obtain
$$
\boxed{U(\beta)U(\alpha)|+\rangle
=e^{-i\alpha/2}\left(\cos(\alpha/2)|+\rangle+i e^{-i\beta}\sin(\alpha/2)|-\rangle\right).}
$$
For example, its <quantum measurement in the computational basis> has probabilities $\mathbb P(k=0)=(1+\sin\alpha\sin\beta)/2$ and $\mathbb P(k=1)=(1-\sin\alpha\sin\beta)/2$, obtained by squaring the two basis amplitudes.
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