I really appreciate how well you have presented your problem. 
People often leave brackets out with questions like these and they become very ambiguous.
I won't claim that i have done it the easiest way. This was a difficult one. But the answer is correct.
4^(0.5log[4](9) - 0.25log[2](25))
$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\
=4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\
=4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\
=4^{log_{4}\;3-log_2\;5^{0.5}}\\\\
=4^{log_{4}\;3-0.5log_2\;5}\\\\$$
Now, I can't do this unless I can get the bases the same.
$$\begin{array}{rll}
let\;\; y&=&log_2 5\\\\
5&=&2^y\\\\
5&=&4^{0.5y}\\\\
log_4 5&=&log_4 4^{0.5y}\\\\
log_4 5&=&0.5ylog_4 4\\\\
log_4 5&=&0.5y\\\\
y&=&2log_4 5\\\\
log_2 5&=&2log_4 5\\\\
\end{array}$$
-------------------------------
so
$$=4^{log_{4}\;3-0.5log_2\;5}\\\\
=4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\
=4^{log_{4}\;3-log_4\;5}\\\\
=4^{log_{4}\;(3/5)}\\\\
=\frac{3}{5}$$
calculator check - using the web2 site calculator.
$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$
The calc has a little rounding error - the answers are the same.
+118723 
