Given:

$\overrightarrow{a}=4\hat{i}+3\hat{j}$

$\overrightarrow{b}=3\hat{i}-4\hat{j}+5\hat{k}$

So,

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 3& 0\\ 3& -4& 5\end{array}\right|$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=15\hat{i}-20\hat{j}-25\hat{k}$

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$

Then,

$\overrightarrow{c}·(\overrightarrow{a}\times \overrightarrow{b})+25=0$

$\Rightarrow 15x-20y-25z+25=0$

$\Rightarrow 3x-4y-5z=-5 ...\left(1\right)$

Also,

$\overrightarrow{c}·\left(\hat{\mathit{i}}\mathit{+}\hat{\mathit{j}}\mathit{+}\hat{\mathit{k}}\right)=4$

$\Rightarrow x+y+z=4 ....\left(2\right)$

And projection of $\overrightarrow{c}$ on $\overrightarrow{a}$ is

$\frac{\overrightarrow{c}·\overrightarrow{a}}{\left|\overrightarrow{a}\right|}=1$

$\Rightarrow \frac{\left(x\hat{i}+y\hat{j}+z\hat{k}\right)·\left(4\hat{i}+3\hat{j}\right)}{\sqrt{16+9}}=1$

$\Rightarrow 4x+3y=5 ....\left(3\right)$

Solving $\left(1\right), \left(2\right) \& \left(3\right)$, we get

$x=2, y=-1, z=3$

So,

$\overrightarrow{c}=2\hat{i}-\hat{j}+3\hat{k}$

Projection of $\overrightarrow{c}$ on $\overrightarrow{b}$ is

$\frac{\overrightarrow{c}·\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{\left(2\hat{i}-\hat{j}+3\hat{k}\right)·\left(3\hat{i}-4\hat{j}+5\hat{k}\right)}{\sqrt{9+16+25}}=\frac{25}{5\sqrt{2}}=\frac{5}{\sqrt{2}}$