Observe that :

$10^{2}$ = 100 and $log_{10}100 = 2$

Let the unknown be y

So we can rewrite $4^{ab+1}$ as

$log_{4}y$ = ab+1

Plugin the values of a and b

$log_{4}y$ = $log_{16}49 \times log_{7}2.5 + 1$

Note that 1 = log_{number}number

Thus we can say that :

$log_{4}y$ = $log_{16}49 \times log_{7}2.5 + log_{4}4$

We can also rewrite $log_{a}b$ into log b / log a

Thus we say that

$log_{4}y$ = ( log 49 / log 16) $\times$ (log 2.5 / log 7) + (log 4 / log 4)

Then you'll see that

$log_{4}y$ = ( log 49 / log 7) $\times$ (log 2.5 / log 16) + (log 4 / log 4) note:swap denominator since its multiplication

$log_{4}y$ = 2 $\times$ (log 2.5 / log 16) + (log 4 / log 4) note: ( log 49 / log 7) = 2, since $7^{2}$ = 49

$log_{4}y$ = 2 $\times$ (log 2.5 / 2log 4) + (log 4 / log 4) note: rules for exponent on logarithms, $log4^{2}$ = 2log4

$log_{4}y$ = (log 2.5 / log 4) + (log 4 / log 4)

$log_{4}y$ = (log 2.5 + log 4 / log 4)

$log_{4}y$ = (log 10 / log 4)

$log_{4}y$ = $log_{4}10$

y = 10

$\log _{ 16 }{ 49\quad =a\quad means\quad { 16 }^{ a }=49\quad }$

$\log _{ 7 }{ 2.5=b\quad means\quad { 7 }^{ b }=2.5 }$

$\quad b=\quad \frac { 1 }{ 2 }$

$since\quad 49\quad \approx \quad { 2.5 }^{ 4 }$

${ 16 }^{ a }={ 2.5 }^{ 4 }$

$\because \quad 16\quad =\quad { 2.5 }^{ 3 }\\ \therefore \quad { 2.5 }^{ 3a }={ 2.5 }^{ 4 }$

$\quad a=\frac { 4 }{ 3 }$

$\therefore \quad { 4 }^{ \frac { 4 }{ 3 } \times \frac { 1 }{ 2 } +1 }=\boxed { 10 }$