Answer
a) ${\bf 0.505}\;\rm \Omega$
b) ${\bf 0.50}\;\rm \Omega$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, from the given graph, that the ammeter and the resistance $R$ are connected in parallel, so they have the same potential difference,
$$V_R=V_A$$
$$I_RR=I_AR_A\tag 1$$
Also we can see that
$$I=I_R+I_A$$
Hence,
$$I_R=I-I_A$$
Plug into (1),
$$(I-I_A)R=I_AR_A\tag 2$$
where we need to make the maximum current of $(\rm 500\;\mu A=0.5\; mA)$ flow through the ammeter while the rest of the current flows through the resistor.
Solving (2) for $R$,
$$R=\dfrac{I_AR_A}{(I-I_A)} $$
Plug the known;
$$R=\dfrac{(0.5 \times 10^{-3})(50)}{(50\times 10^{-3})-(0.5 \times 10^{-3})} $$
$$R=\color{red}{\bf 0.505}\;\rm \Omega$$
$$\color{blue}{\bf [b]}$$
The effective resistance of the ammeter is given by
$$R_{eq}=\left[ R^{-1}+R_{A}^{-1} \right]^{-1}$$
Plug the known;
$$R_{eq}=\left[ 0.505^{-1}+50^{-1} \right]^{-1}$$
$$R_{eq}=\color{red}{\bf 0.50}\;\rm \Omega$$
