(i) We can show that any element $(x,y,z)\in \mathbb R^3$ can be written as the sum of two elements in $U$ and $V$. Any such element is of the form
$$(a,a-b,a+b)+(0,0,c)=(a,a-b,a+b+c)$$
for some $a,b,c\in \mathbb R$. So, we can pick $a=x$, $b=x-y$ and $c=-(2x-y)+z$. Then $(x,y,z)=(a,a-b,a+b+c)$ and thus $U+V=\mathbb R^3$. You have already shown that $U\cap V=\{(0,0,0)\}$ (if $(a,a-b,a+b)=(0,0,c)$, then $a=b=0$ and hence $c=0$), so the sum is direct.
(ii) This can be done similarly to (i), but I'll illustrate a slightly different method. The sum of any $u\in U$ and $w\in W$ is of the form
$$u+w=(a,a-b,a+b)+(d,e,e)=(a+d,a-b+e,a+b+e)=(f+d-e,f-b,f+b) ~~~~\text{where $f=a+e.$}$$
Then $u+w=f(1,1,1)+d(1,0,0)+e(0,-1,1)$ for some $f,d,e\in\mathbb R$. Therefore $U+W=\mathrm{Span}(A)$, where $A=\{(1,1,1),(1,0,0),(0,-1,1)\}$. We'll show that $A$ is linearly independent in $\mathbb R^3$. If $f(1,1,1)+d(1,0,0)+e(0,-1,1)=0$ for some $f,d,e\in \mathbb R$ then
$$(i)~f+d=0,~~~~~(ii)~f-e=0,~~~~~(iii)~f+e=0.$$
Sum $(ii)$ and $(iii)$ to obtain $f=0$ and substract the same equations to obtain $e=0$. Replacing in $(i)$ we get $d=0$ and thus $A$ is a linearly independent set in $\mathbb R^3$. However, $\dim_{\mathbb R} \mathbb R^3=3$, so $\mathbb R^3 = \mathrm{Span}(A)=U+W$ since $A$ is a linearly independent subset of $\mathbb R^3$ with $3$ elements.
Best Answer
Note that if $S=\text{span}\left\{\begin{bmatrix} -3\\ 4\\ 1 \end{bmatrix}\right\}$, then $S=\left\{a\begin{bmatrix} -3\\ 4\\ 1 \end{bmatrix}:a\in\mathbb{R}\right\}$. This vector space is isomorphic to $\tilde{S}=\left\{a(-3,4,1):a\in\mathbb{R}\right\}$.
Also, note that $T=\left\{(x,y,z)\in\mathbb{R}^3:kx+y-z=0\right\}\Leftrightarrow T=\{(x,y,kx+y):x,y\in\mathbb{R}\}$.
Since $\mathbb{R}^3=\tilde{S}\oplus T$, we must have $\tilde{S}\cap T=\{(0,0,0)\}$. This occurs for any real number $k\neq 1$ (why?).