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.

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"This shows that $D=0$." This isn't true. You can have divisors numerically equivalent to $0$, but which are not linearly equivalent to $0$. For example, consider the canonical divisor on an Enriques surface. –  Matt Nov 25 '12 at 17:47
@Matt Thanks for the response. I will update my working if I manage to think of a new solution. Please feel free to make any more comments. –  Yong Hao Ng Nov 25 '12 at 18:10

(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.

(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.
May I ask why "deg$D'=D'.L$" and "div$(f).L=0$"? Namely, if degree of divisor is $d$, does it work like a degree $d$ curve and hence Bezout's theorem applies? –  Yong Hao Ng Nov 26 '12 at 2:52
It is defined as $\sum n_i$ where $D=\sum n_iD_i$ and $D_i$ are prime divisors. –  Yong Hao Ng Nov 26 '12 at 10:29
$D_i$ are points on $X$ and $D$ is a divisor on $X$. –  Yong Hao Ng Nov 26 '12 at 10:34
OK (though $D_i$ should not be a point !). So just say $D'.L>0$ (take a line $L$ which doesn' contain any irreducible component of $D'$. For the second equality, write $f$ as quotient $P/Q$ of two homogeneous polynomials of same degree $d$. Then $\mathrm{div}(f)=[Z(P)]-[Z(Q)]$, so $\mathrm{div}(f).L=[Z(P)].L-[Z(Q)].L=d-d=0$ by Bézout. –  user18119 Nov 26 '12 at 10:40