I have two functions, $f:\mathbb{Q}\to\mathbb R_+$ given by $f(x) = \frac{1}{5} \sin{\left (\pi x \right )} + 4 $, and $g:\mathbb{Z_+}\to\mathbb{Q}$ given by $g(x) = - \frac{7 x}{2}$. I needed to find the composite function $h(x)=f(g(x))$
and I got the following result: $\frac{1}{5}\sin \left(-\frac{7\pi x}{2}\right)+4$
I know that the function above is not injective since h[1] = h[5] but why isn't this function surjective?
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Hint. For all $x\in V\subset\mathbb{R}$ it holds that $$-1 \leq \sin(x)\leq 1$$ Hence we have that $$-\frac15+4\leq \frac15\sin(\pi g(x)) + 4\leq \frac15+4$$ and therefore $$\frac{19}{5}\leq f(g(x))\leq \frac{21}{5}$$
Now consider any value outside that range.
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We have $h : \mathbb{Z}_+ \to \mathbb{R}_+$, a surjective function $h$ would for every $y \in \mathbb{R}_+$ have a $x \in \mathbb{Z}_+$ with $y = h(x)$.
This would mean that one could find an enumeration for $\mathbb{R}_+$, a surjective function $\varphi : \mathbb{N} \to \mathbb{R}_+$, which is not possible.
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It's not surjective because there are real numbers that you can't get as a result of this function. $\sin(y)$ is between $-1$ and $1$, so you can't get any result outside the range $[4-\frac15,4+\frac 15]$. In fact you can't get most results inside that range either, since $h(n+4)=h(n)$ means that it actually only takes four different values.
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It's because $\mathbb Q$ is countable while $\mathbb R$ isn't. Apart from the fact that the function is bounded you have that you misses values within it's bounds too, consider the equation:
$${1\over 5}\sin{-7\pi x\over 2} + 4 = {1\over 5}\sin C +4$$
has the solutions $-7\pi x/2 = \pi/2 \pm (\pi/2 - C)$ that is:
$$x = -{2\over 7 \pi} (\pi/2 \pm (\pi/2 - C) = -{1 \pm ( - 2C/\pi)\over 7 }$$
Now if $2C/\pi$ is irrational (for example $1/\sqrt{2}$) the solutions would be all irrational. This means the function don't take those values either.
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Whether it is surjective or not depends on the codomain. You can define a codomain as a set of all possible real outputs and thus make the function surjective.
However, assuming most naturally the codomain is $\mathbb R$ (because we talk about real-valued functions) or $\mathbb R_+$ (because that's a codomain of $f$, which is external in the requested function composition), the function $h$ obtained is not surjective, because the sine function is bounded by $+1$, so the maximum value of $\frac 15\,\sin(anything)+4$ will not exceed $\frac{21}5$, hence $h$ values will not cover the range $(\frac{21}5,\infty)$, which is a subset of the codomain (either $\mathbb R$ or $\mathbb R_+$).
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When you noticed $h(1)=h(5)$ (don't use square brackets for this), you may have noticed that actually $h(x+4) = h(x)$ for all $x$. Therefore the image of $h$ consists of $h(1)$, $h(2)$, $h(3)$, and $h(4)$ only. Since these are only finitely many values, they are not all of $\mathbb R_+$.
Indeed, you also have $h(2)=4=h(4)$, the image consists of three numbers.
It's not surjective because it's a function between $\mathbb Z_+$ and $\mathbb R_+$. We know that $|\mathbb Z_+| < |\mathbb R_+|$, so there can't be any surjection from the smaller set to the larger.
Another way to actually show that your function is not surjective is to show that $f(g(x))\leq \frac92$, therefore there is no element $x\in\mathbb Z_+$ such that $f(g(x))=10$.