Q:

Cartier divisor and Dimension of Cohomology Group

I am doing some practice questions for my exam and I would appreciate help in solving this problem:

$D,E$ are Cartier divisors on a nonsingular projective surface $X$.
  (1) If $D\equiv 0$ show that dim $H^0(X,sD)\leq 1$ for all $s\geq 1$.
  (2) If $D$ is effective, $D\neq 0$ and $D\equiv E$ show that $H^0(X,tE)=0$ for all $t\leq -1$.
  (3) Is (2) true if $E=D$ and $X$ is nonsingular affine variety of positive dimension? 

Here is my workings for (1) and (2): are they correct? 
By definition, $D\equiv 0$ means $D.C=0.C=0$ for any curve $C$. 
(Note: The remainder of this portion was pointed out to be wrong)
Since I can easily construct a line intersecting with $D$, this shows that $D=0$.   Then clearly $sD=0$ for all $s\geq 1$.   I know that dim $H^0(X,sD)-1=$ dim$|sD|$, so I can try to show that dim $|sD|\leq 0$.  Since $|sD|=\lbrace D'\geq 0|D'\sim sD\rbrace$, therefore $D'\sim 0$ and $D'=0$.   This shows that $|sD|=\lbrace 0\rbrace$ and hence dim $|sD|=0$.
For (2), since $D>0$ and $t\geq -1$, hence deg $tE<0$.
$H^0(X,tE)=\lbrace f\in k(X)\setminus\lbrace 0\rbrace|tE+div(f)\geq 0\rbrace\cup\lbrace 0\rbrace$.
Assuming such an $f$ exists, then $tE+div(f)\geq 0$.
But degree of $div(f)=0$, hence this is a contradiction.   
Not sure about (3)...
Thanks for reading! 

A:

(1) If $h^0(sD)\ge 1$, then there exists a rational function $f$ such that $\mathrm{div}(f)+sD=D'\ge 0$. If $D'=0$, then $sD$ is principal and $h^0=1$. Suppose $D'>0$. Let $L$ be a line, then 
$$0<\deg D'=D'. L=\mathrm{div}(f).L+sD.L=0,$$
contradiction.  
(2) Your proof is correct. 
(3) On an affine variety $V$, a coherent sheaf $\mathscr F$ is zero if and only if $H^0(V, \mathscr F)=0$. As $O_X(tE)$ is an invertible sheaf, it is never zero unless $X$ is empty.  So $H^0(X, tE)\ne 0$ if $X$ is affine and non-empty.