My Try:

# $$\Z_{in}=Z_0*\frac{Z_L+jZ_0tan\beta l}{Z_0+jZ_L tan\beta l}\$$

Now

$$\\beta l=\frac{2\pi\times37.5\times10^6\times10}{3\times10^8}=\frac{5\pi}{2}\$$

so its becomes

$$\Z_{in}=\frac{Z_0^2}{Z_L} \$$

$$\Z_{in}=\frac{200\times200}{100}=400\Omega\$$

So for part $$\(a)\$$

$$\I_{i}=\frac{200}{600}=\frac{1}{3}A\$$ also $$\V_{i}=\frac{400\times200}{600}=\frac{400}{3}\$$

Now for part $$\(b)\$$ from the generator we know the equation for lossless line is $$\ I(z)=\frac{-j}{Z_0}sin\beta z V_i+Cos\beta z I_{i}\$$

so current at the load, that means at $$\\lambda/4\$$ distance from the generator will be $$\I=\frac{-j}{200}sin(\frac{5\pi}{2})\times\frac{400}{3}+0\times\frac{1}{3}=\frac{2}{3}\angle-90^{\circ}\$$

but for part$$\(b)\$$ the actual answer is $$\\frac{1}{3}\angle-90^{\circ}\$$

What is the mistake i am doing can anyone help please?

For these problems, remember that the $z=0$ location (highlighted in the image below) is defined at the terminals of the load. So you are looking for $I(0)$ in part b.
For the input impedance, you do use the length because it is some distance $l$ away from the load and towards the generator. This is $z=-l$.