= Solution
Both candidate sets contain zero. The <sum of subspaces> is closed under <linear combinations> because $a(u_1+w_1)+b(u_2+w_2)=(au_1+bu_2)+(aw_1+bw_2)$, with the two terms in $U$ and $W$. The intersection is closed because any <linear combination> of vectors lying in both subspaces still lies in both. Thus each is a <vector subspace>.
Take a <basis> $s_1,\ldots,s_r$ of the intersection, extend it by $u_1,\ldots,u_a$ to a <basis> of $U$, and by $w_1,\ldots,w_b$ to a <basis> of $W$. The combined list $s,u,w$ spans $U+W$. To prove independence, a zero combination makes the $u$ combination equal a vector in both $U$ and $W$, so it is a combination of the $s$'s. Independence of the <basis> of $U$ makes all $u$ coefficients zero; independence of the <basis> of $W$ then makes all remaining coefficients zero. Hence $\dim(U+W)=r+a+b$, giving the <dimension formula for a sum of subspaces>
$$
\boxed{\dim U+\dim W=\dim(U+W)+\dim(U\cap W).}
$$
For the concrete kernels, the equations $Ax=0$ give
$$
x=\left(\frac{5s-5t}{3},\frac{-s+7t}{3},s,t\right).
$$
Applying the two rows of $B$ to this expression gives $4(s-t)$ and $5(s-t)/3$. Thus the intersection has <basis> $v_0=(0,2,1,1)$. Choose $v_1=(5,-1,3,0)$ in $U$ and $v_2=(-4,-2,1,0)$ in $W$. Then
$$
\boxed{\mathcal B_{U\cap W}=(v_0),\quad\mathcal B_U=(v_0,v_1),\quad
\mathcal B_{U+W}=(v_0,v_1,v_2).}
$$
Indeed $U$ has dimension two because $A$ has rank two, and $v_0,v_1$ are independent. Likewise $W$ has dimension two and $v_0,v_2$ are independent, so the last list spans the sum. Its first three coordinates have determinant $-48$, confirming independence and dimension three.
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