Can we move curves which are members of very ample systems? Let us take the second degree Hirzebruch surface $\mathbb{F}_2$ which is a holomorphic $\mathbb{CP}^1$ bundle over $\mathbb{CP}^1$ having sections of self intersections $+2$ and $-2$. Let me denote the class of the $-2$ section by $C$ and class of the sphere fiber by $F$. The Picard group of $\mathbb{F}_2$ is generated by $C$ and $F$.

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*It is known that the linear system $|2C+5F|$ is very ample and it contains an irreducible curve, let me denote it by $2C+5F$.
On the other hand, I take the linear system $|C+2F|$. By the same proposition in Hartshorne, this system is not very ample but contains an irreducible curve, denoted by $C+2F$.

The curves $(2C+5F)$ and $(C+2F)$ intersect at $5$ points generically. My question is: Can we move $2C+5F$ so that it intersects $C+2F$ at one point with multiplicity 5$$? But while moving I do not want to change the genera and self intersections. Or let me put it another way; is there a member of the very ample system $|2C+5F|$ which intersects a member of $|C+2F|$ at one point with multiplicity $5$?
As I know in very ample systems there is more freedom, since they are not rigid. So I was hoping the curve $2C+5F$ in the way that I want, but I could not find any proof in relation to these.


*My second question: The fiber $F$ and the $-2$ section $C$ of $\mathbb{F}_2$ intersect once, say, at the point $p$. Can we find a member of the very ample linear system $|2C+5F|$ which passes through this point $p$ such that, at $p$,  $C$ and $2C+5F$ intersect once, and $F$ and $2C+5F$ intersect once but with multiplicity $2$?

I would appreciate any suggestions or resources. Thanks.
 A: I am just writing up my comment as an answer.  For every smooth, genus $2$ curve $X$, the image of the Abel map, $$u_1:X\to \text{Pic}^1(X), \ \ p\mapsto [\mathcal{O}_X(\underline{p})],$$ is a Cartier divisor in the algebraic equivalence class of the theta divisor $\theta$.  Similarly, for every integer $n\neq 0$, the composite morphism $u_n$, $$X\to \text{Pic}^1(X) \xrightarrow{m_n} \text{Pic}^n(X), \ \ p\mapsto [\mathcal{O}_X(n\underline{p})],$$ is a Cartier divisor in the algebraic equivalence class of $|n|\cdot \theta.$  In particular, for every Cartier divisor class $D$ on $X$ of degree $4$, the two divisor classes $D+[u_1(X)]$ and $[u_5(X)]$ on $\text{Pic}^5(X)$ are $\theta$ and $5\theta$.  Thus, the intersection number of these divisor classes is $$(D+[u_1(X)])\cdot [u_5(X)] = 5\theta^2 = 10[\{*\}].$$  Thus, for every degree $4$ divisor class $D$, when counted with the appropriate multiplicities, the following scheme has length $10$, $$Z_D:=\{(p,q)\in X\times X | 5\underline{q} \sim \underline{p} + D \}.$$
When $D$ is the divisor class $2K_X$, then $Z_D$ contains $(r,r)$ for each of the $6$ Weierstrass points $r$ of $X$.  Since $10>6$, there exists $(p,q)\in Z_D$ with $p\neq q$.  Now consider the complete linear system of the divisor class $\underline{p} + 2K_X$.  By Riemann-Roch and Serre duality, this complete linear system is a $g^3_5$ on $X$ that embeds $X$ as a degree $5$ curves in $\mathbb{P}^3$.  
Moreover, the composition of this embedding with linear projection away from $p$ is a $g^2_4$ on $X$ given by the complete linear system of $2K_X$, i.e., it is the composition of the canonical $g^1_2$ from $X$ to $\mathbb{P}^1$ and the Veronese $2$-uple map from $\mathbb{P}^1$ to a smooth conic $\overline{Q}$ in $\mathbb{P}^2$.  Since linear projection away from $p$ factors through the conic, the image of $X$ in $\mathbb{P}^3$ is contained in the singular quadric cone $Q$ over the smooth plane conic $\overline{Q}$.  
Of course the minimal desingularization of $Q$ is a Hirzebruch surface $\mathbb{F}_2$.  Also, the divisor class of the image of $X$ in $\mathbb{F}_2$ equals $2C+5F$.  Finally, by construction, the osculating hyperplane $H_q$ to $X$ in $\mathbb{P}^3$ at $q$ has contact order $5$ with $X$.  The hyperplane section $H_q\cap Q$ is a plane conic contained in $Q$ whose divisor class in $\mathbb{F}_2$ equals $C+2F$.  Since $q$ does not equal $p$, and since $X$ does contain $p$, the hyperplane section is not reducible (every reducible hyperplane section of $Q$ contains the vertex $p$ of the cone).  Thus, the hyperplane section is a smooth curve $D$.  Altogether, the strict transforms of $X$ and $D$ in $\mathbb{F}_2$ are smooth curves with respective divisor classes $2C+5F$ and $C+2F$ such that $X\cdot D$ equals $q$ with multiplicity $5$.
The answer to the second question is similar and easier: just use the linear system $\underline{r}+2K_X$ on $X$ where $r$ is one of the $6$ Weierstrass points of $X$.
