For simplicity, considering log with base $2$.
$f_1 = 2^{\lg n}\; =\; n^{\lg 2}$
$f_2= n^{10}$
So, $f_1$ has slower growth rate than $f_2$ i.e.
--- $f_1 << f_2$
$f_3 = (\sqrt{\lg n})^{\lg^2n} =(\lg n)^{\frac{\lg ^{2}n}{2}} = (\lg n)^{\frac{(\lg n)^2}{2}}$
$f_4 = (\lg n)^{\sqrt{\lg n}} = (\lg n)^{(\lg n)^{\frac{1}{2}}}$
$\because$ $(\lg n)^{\frac{1}{2}} < \frac{(\lg n)^2}{2}.$ (For large values of $'n'$). So,
--- $f_4 << f_3.$
Between $f_1$ and $f_4 :$
$f_4$ can be rewritten as $2^{\lg \; ((\lg n)^{\sqrt{\lg n}})}= 2^{\sqrt{\lg n}\;\lg \lg n}$
So, comparison is between $2^{\lg n}$ and $2^{\sqrt{\lg n}\;\lg \lg n}$
$\because$ $\lg n > \sqrt{\lg n}\;\lg \lg n$ ( $\sqrt{\lg n}$ is common in both functions). So,
--- $f_4 << f_1.$
Till now, among $f_1,f_2,f_3,f_4$, slowest function is $f_4$. If we get $f_5 << f_4$ then $f_5$ will be slowest otherwise $f_4.$
Between $f_4$ and $f_5 :$
$f_5 = 2^{2^{\sqrt{lglgn}}} $
$f_4$ can also be rewritten as $2^{2^{\lg (\sqrt{\lg n}\;\lg \lg n)})}$
Now, $\lg (\sqrt{\lg n}\;\lg \lg n) =\lg \sqrt{\lg n} + \lg \lg \lg n = \frac{1}{2} \lg \lg n + \lg \lg \lg n > \sqrt{\lg \lg n}.$
-- $f_5 << f_4.$
Hence, function $f_5$ grows slowest asymptotically.
Now, to get the correct order, we have to compare $f_2$ and $f_3.$
Between $f_2$ and $f_3 :$
set $n=2^m$
$f_2 = (2^{m})^{10}$
$f_3 = m^{\frac{m^2}{2}} = (m^m)^{m/2}$
Since, $m^m >> m! >> 2^m.$ So,
-- $f_2 << f_3.$
Or, we can do like this :
$f_2 = n^{10} = 2^ {\lg n^{10}} = 2^{10 \lg n}$
$f_3 = (\lg n)^{\frac{(\lg n)^2}{2}} = 2^{\lg ( (\lg n)^{\frac{(\lg n)^2}{2}})} = 2^{\frac{(\lg n)^2}{2} \lg \lg n}$
$\because 10 \lg n < \frac{(\lg n)^2}{2} \lg \lg n.$ So, $f_2 << f_3.$
Therefore, increasing order of growth rate is :
$f_5 << f_4 << f_1 << f_2 << f_3$