I agree that this question is rather elementary for this site but since Easter is around the corner...
Assume the $X_i$'s are independent. Suppose that $X_i$ takes the values $a_i$ and $ b_i $, with probabilities $p_i(a_i)$ and respectively $p_i(b_i)$, where $p_i(a_i)+p_1(b_i)=1$. Without loss of generality we can assume $a_i< b_i$.
The vector valued r.v. $\vec{X}=(X_1,\dotsc, X_n)$ is distributed on the set $V$ of vertices of the parallelepiped
$$ P=\prod_{i=1}^n [a_i,b_i]. $$
A vertex $\vec{v}$ of this parallelepiped has coordinates
$$ \vec{v}=(v_1,\dotsc, v_n),\;\;v_i\in\lbrace a_i,b_i\rbrace. $$
The probability that $\vec{X}=\vec{v}$ is
$$p(\vec{v})=\prod_{i=1}^n p_i(v_i). $$
The distribution $\mu$ of $f(\vec{X})$ is a sum of delta measures
$$ \mu=\sum_{t\in \mathbb{R}} w_t \delta_t, $$
where
$$w_t =\sum_{f(\vec{v})=t} p(\vec{v}). $$
In the end the problems reduces to identifying which of the vertices of $V$ lies on a given level set of $f$ which may not be easy for a complicated $f$. If $a_1=\cdots =a_n=a$, $b_1=\cdots =b_n=b$, $p_1(a)=\cdots =p_n(a)=p$ and $p_1(b)=\cdots = p_n(b)=q=1-p$ the above formula simplifies somewhat.