For the pedantic's sake, we first have a polynomial reduction $3SAT \leq_p s3SAT$, where the later has strictly 3 terms per clause, not less (as accepted by the former). This is achieved by taking an instance of $3SAT$ and mapping the clauses respectively:
$ (a) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor s \lor a) $
$ (a \lor b) \mapsto (\lnot s \lor \lnot s \lor \lnot s) \land (s \lor b \lor a) $
where $s$ is a dummy variable, and a and b are terms where $\lnot \lnot$ is negated respectively. We concatenate the end result with $\land$. Clearly we construct in polynomial time.
Next we do the polynomial reduction $s3SAT \leq_p NAE-4SAT$, where the second is simply $NAE-3SAT$ but with four terms per clause. For this we take an instance of $s3SAT$ and map each clause as follows:
$ (x \lor y \lor z) \mapsto (x \lor y \lor z \lor s) \land (\lnot x \lor \lnot y \lor \lnot z \lor \lnot s)$
Where $x,y,z$ are terms and $s$ is our dummy variable, as before. Notice the symmetry here, if we find an assignment with $s = true$ we can simply invert the assignment to receive another valid assignment with $s=false$. This is the assignment we want.
As a last step we reduce $NAE-4SAT \leq_p NAE-3SAT$. Take an instance in $NAE-4SAT$ and map the clauses as follows:
$ (a \lor b \lor c \lor d) \mapsto (s \lor a \lor b) \land (\lnot s \lor c \lor d)$
All the same as before, concatenate the result once more. Notice here that if the true and false variable's (one of each must exist) are mapped to the same clause, $s$ can be choose appropriately. If they are mapped to different clauses, $s$ can be choose opposite to the respective variable value in each pair.
To summarize: $3SAT \leq_p s3-SAT \leq_p NAE_4SAT \leq_p NAE-3SAT$.
Answer to extra question: 3SAT only allows $\lor$ in the clauses.
The reduction is straightforward. As an example, I'll reduce an instance of Exactly-One-in-3SAT to an instance of Unit Intersection.
$$ (x_1 \lor x_4 \lor x_3) \land (\overline{x_4} \lor \overline{x_2} \lor x_3) \land (x_2 \lor x_1 \lor \overline{x_3}) $$
Let $n$ be the number of distinct (positive and negative) literals in the formula, and choose a bijection between the literals and the integers from $1$ to $n$:
$$x_1 ↔ 1\\
x_2 ↔ 2\\
x_3 ↔ 3\\
x_4 ↔ 4\\
\overline{x_2} ↔ 5\\
\overline{x_3} ↔ 6\\
\overline{x_4} ↔ 7$$
The integers on the right are the set $X$ in the Unit Intersection
instance: $X = \lbrace 1,2,3,4,5,6,7 \rbrace$.
Convert each clause in the Exactly-One-in-3SAT instance into a set of integers by using the mapping created earlier to replace literals with integers.
$$ (x_1 \lor x_4 \lor x_3) \rightarrow \lbrace 1, 4, 3 \rbrace \\
(\overline{x_4} \lor \overline{x_2} \lor x_3) \rightarrow \lbrace 7, 5, 3 \rbrace \\
(x_2 \lor x_1 \lor \overline{x_3}) \rightarrow \lbrace 2, 1, 6 \rbrace$$
These are the initial subsets $S_1 ... S_m$ in the Unit Intersection instance.
If variables appear as both positive and negative literals in the Exactly-One-in-3SAT instance, for each such variable add a subset that contains the mapped integers for the positive and negative literals of that variable. For our instance we would add
$$\lbrace 2, 5 \rbrace \\
\lbrace 3, 6 \rbrace \\
\lbrace 4, 7 \rbrace$$
The reduction is complete. If a subset $T$ can be found for the Unit Intersection instance, that solution can be mapped back to a satisfying assignment for the Exactly-One-in-3SAT instance.
Best Answer
Hint: let $f$ be a 3SAT formula and $w$ a variable not contained in $f$. Then $(w \vee w \vee \overline{w}) \wedge f$ has at least two solutions iff $f$ has at least one.