= Solution
Use the column order $1,(123),(12)(34),(12345),(13452)$ in $A_5$. The two restrictions of degree four coincide and equal $\theta_4=(4,1,0,-1,-1)$; those of degree five coincide and equal $\theta_5=(5,-1,1,0,0)$. The <centralizer> orders give
$$
\langle\theta_4,\theta_4\rangle=\frac{16}{60}+\frac13+\frac15+\frac15=1,\qquad
\langle\theta_5,\theta_5\rangle=\frac{25}{60}+\frac13+\frac14=1.
$$
Thus both are <irreducible characters>. The remaining given restriction is $\eta=(6,0,-2,1,1)$, and
$$
\langle\eta,\eta\rangle=\frac{36}{60}+\frac44+\frac15+\frac15=2.
$$
It splits as two distinct <irreducible characters>. Neither is $1$, $\theta_4$ or $\theta_5$, since their <character inner products> with $\eta$ are zero. A finite <group> has as many complex <irreducible characters> as <conjugacy classes>, so these five exhaust them. The sum of squares of irreducible degrees is $|A_5|=60$. The first three degrees contribute $1+16+25=42$, leaving $18$ for the two constituents, whose degrees sum to six. Hence both have degree three.
Write one constituent as $\theta=(3,x,y,u,v)$. All five <conjugacy classes> are closed under inversion, so all its values are real: $\chi(g^{-1})=\overline{\chi(g)}$. Orthogonality with $1$, $\theta_4$ and $\theta_5$ gives
$$
\frac3{60}+\frac x3+\frac y4+\frac{u+v}5=0,\qquad
\frac{12}{60}+\frac x3-\frac{u+v}5=0,\qquad
\frac{15}{60}-\frac x3+\frac y4=0.
$$
Solving gives $x=0$, $y=-1$ and $u+v=1$. The squared <norm> one gives $9/60+1/4+(u^2+v^2)/5=1$, so $u^2+v^2=3$ and $uv=-1$. Thus $u,v$ are the roots of $t^2-t-1$, namely $\varphi=(1+\sqrt5)/2$ and $\bar\varphi=(1-\sqrt5)/2$. Here the bar labels the other real root, rather than complex conjugation. The other constituent is $\eta-\theta$, so the two last values are interchanged.
The <Character table of the alternating group A5> is therefore
$$
\boxed{\begin{array}{c|rrrrr}
&1&(123)&(12)(34)&(12345)&(13452)\\\hline
1&1&1&1&1&1\\
\theta_3&3&0&-1&\varphi&\bar\varphi\\
\theta'_3&3&0&-1&\bar\varphi&\varphi\\
\theta_4&4&1&0&-1&-1\\
\theta_5&5&-1&1&0&0
\end{array}.}
$$
The <conjugacy class> sizes are $1,20,15,12,12$. A <normal subgroup> is a union of entire <conjugacy classes> containing the identity, and its order divides $60$ by <Lagrange's theorem>. A nontrivial such union has size at least $13$; the only proper divisors of $60$ at least $13$ are $15,20,30$. Subtracting the identity would require a sum of some of $20,15,12,12$ equal to $14,19,29$, respectively, none of which is possible. Thus \b[$A_5$ is simple].
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