= Solution
Let $r=dim(U\cap W)$. Choose a <basis> $e_1,\ldots,e_r$ of the <intersection of vector subspaces>, extend it to a basis $e_1,\ldots,e_r,u_{r+1},\ldots,u_m$ of $U$, and independently extend it to a basis $e_1,\ldots,e_r,w_{r+1},\ldots,w_n$ of $W$. The combined list
$$
e_1,\ldots,e_r,u_{r+1},\ldots,u_m,w_{r+1},\ldots,w_n
$$
spans the <sum of vector subspaces> $U+W$. It is also <linear independence>[linearly independent]: a vanishing <linear combination> says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the <dimension formula for a sum of subspaces>
$$
\boxed{\dim(U+W)=\dim U+\dim W-\dim(U\cap W)}.
$$
Parametrize the two <linear subspaces> as
$$
U=\operatorname{span}\{(7,-5,1,0),(8,-6,0,1)\},
\qquad
W=\operatorname{span}\{(-2,1,0,0),(-3,0,1,0)\}.
$$
Putting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thus
$$
U\cap W=\operatorname{span}\{(7,-5,1,0)\},
$$
and the dimension formula gives $\dim(U+W)=2+2-1=3$.
The <linear functional>
$$
\boxed{\ell(x)=x_1+2x_2+3x_3+4x_4}
$$
vanishes on each displayed generator of $U$ and $W$. Its <kernel> is three-dimensional because $\ell\ne0$, so the inclusion $U+W\subseteq\ker\ell$ between two three-dimensional subspaces is equality.
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